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Integration

1.1 Definition of the Integral

47 problems · hints, answers and solutions shown beside each one

1Stage 1Conceptual

Do you understand the idea? Usually little or no calculation.

For Questions 1 through 5, we want you to develop an understanding of the model we are using to define an integral: we approximate the area under a curve by bounding it between rectangles. Later, we will learn more sophisticated methods of integration, but they are all based on this simple concept.

Q1Stage 1

Give a range of possible values for the shaded area in the picture below.

Figure from prob_s1.1, line 2

Figure from prob_s1.1, line 2

Hint

Draw a rectangle that encompasses the entire shaded area, and one that is encompassed by the shaded area. The shaded area is no more than the area of the bigger rectangle, and no less than the area of the smaller rectangle.

Answer

The area is between 1.51.5 and 2.52.5 square units.

Full solution

Figure from prob_s1.1, line 2

Figure from prob_s1.1, line 2

Figure from prob_s1.1, line 13

Figure from prob_s1.1, line 13

The diagram on the left shows a rectangle with area 2×1.25=2.52 \times 1.25=2.5 square units. Since the blue-shaded region is entirely inside this rectangle, the area of the blue-shaded region is no more than 2.5 square units.

The diagram on the right shows a rectangle with area 2×0.75=1.52 \times 0.75=1.5 square units. Since the blue-shaded region contains this entire rectangle, the area of the blue region is no less than 1.5 square units.

So, the area of the blue-shaded region is between 1.5 and 2.5 square units.

Remark: we could also give an obvious range, like “the shaded area is between zero and one million square units." This would be true, but not very useful or interesting.

Q2Stage 1

Give a range of possible values for the shaded area in the picture below.

Figure from prob_s1.1, line 2

Figure from prob_s1.1, line 2

Hint

We can improve on the method of Question 1 by using three rectangles that together encompass the shaded region, and three rectangles that together are encompassed by the shaded region.

Answer

The shaded area is between 2.75 and 4.25 square units. (Other estimates are possible, but this is a reasonable estimate, using methods from this chapter.)

Full solution

  • One naive way to solve this is to simply use the same method as Question 1.

    Figure from prob_s1.1, line 2

    Figure from prob_s1.1, line 2

    Figure from prob_s1.1, line 21

    Figure from prob_s1.1, line 21

    The rectangle on the left has area 3×2.25=6.753 \times 2.25 = 6.75 square units, and encompasses the entire shaded region. The rectangle on the right has area 3×0.25=0.753 \times 0.25 = 0.75 square units, and is entirely contained inside the blue-shaded region. So, the area of the blue-shaded region is between 0.75 and 6.75 square units.

    This is a legitimate approximation, but we can easily do much better. The shape of this graph suggests that using the areas of three rectangles would be a natural way to improve our estimate.

  • Let's use these rectangles instead:

    Figure from prob_s1.1, line 2

    Figure from prob_s1.1, line 2

    Figure from prob_s1.1, line 26

    Figure from prob_s1.1, line 26

    In the left picture, the red area is (1×1.25)+(1×2.25)+(1×0.75)=4.25(1 \times 1.25)+(1 \times 2.25)+(1 \times 0.75)=4.25 square units. In the right picture, the red area is (1×0.75)+(1×1.75)+(1×0.25)=2.75(1 \times 0.75)+(1 \times 1.75)+(1 \times 0.25)=2.75 square units. So, the blue shaded area is between 2.75 and 4.25 square units.

Q3Stage 1

Using rectangles, find a lower and upper bound for ∫1312xdx\displaystyle\int_1^3 \dfrac{1}{2^x}\dee{x} that differ by at most 0.2 square units.

Figure from prob_s1.1, line 2

Figure from prob_s1.1, line 2

Hint

Four rectangles suffice.

Answer

The area under the curve is a number in the interval (38[12+12],38[1+12])\left( \frac{3}{8}\left[\frac{1}{2}+\frac{1}{\sqrt{2}}\right], \frac{3}{8}\left[1+\frac{1}{\sqrt{2}}\right]\right).

Full solution

Remark: in the solution below, we find the appropriate approximation using trial and error. In Question 46, we take a more systematic approach.

  • First, we can try by using a single rectangle as an overestimate, and a single rectangle as an underestimate.

    Figure from prob_s1.1, line 2

    Figure from prob_s1.1, line 2

    Figure from prob_s1.1, line 13

    Figure from prob_s1.1, line 13

    The area under the curve is less than the area of the rectangle on the left (2×12=12 \times \frac{1}{2}=1) and greater than the area of the rectangle on the right (2×18=142 \times \frac{1}{8}=\frac{1}{4}). So, the area is in the range (14,1)\left(\frac{1}{4},1\right). Unfortunately, this range is too big–we need our range to have length at most 0.2. So, we refine our approximation by using more rectangles.

  • Let's try using two rectangles each for the upper and lower bounds.

    Figure from prob_s1.1, line 2

    Figure from prob_s1.1, line 2

    Figure from prob_s1.1, line 15

    Figure from prob_s1.1, line 15

    The rectangles in the left picture have area (1×12)+(1×14)=34\left(1 \times \frac{1}{2}\right)+\left(1 \times \frac{1}{4}\right)=\frac{3}{4}, and the rectangles in the right picture have area (1×14)+(1×18)=38\left(1 \times \frac{1}{4}\right)+\left(1 \times \frac{1}{8}\right)=\frac{3}{8}. So, the area under the curve is in the interval (38,34)\left(\frac{3}{8},\frac{3}{4}\right). The length of this interval is 38\frac{3}{8}, and 38>315=15=0.2\frac{3}{8}>\frac{3}{15}=\frac{1}{5}=0.2. (Indeed, 38=0.375>0.2\frac{3}{8}=0.375>0.2.) Since the length of our interval is still bigger than 0.2, we need even more rectangles.

  • Let's go ahead and try four rectangles each for the upper and lower estimates.

    Figure from prob_s1.1, line 2

    Figure from prob_s1.1, line 2

    Figure from prob_s1.1, line 20

    Figure from prob_s1.1, line 20

    The area of the rectangles on the left is:

    (12×12)+(12×122)+(12×14)+(12×142)=38[1+12],\left(\frac{1}{2}\times \frac{1}{2}\right)+ \left(\frac{1}{2}\times \frac{1}{2\sqrt{2}}\right)+ \left(\frac{1}{2}\times \frac{1}{4}\right)+ \left(\frac{1}{2}\times \frac{1}{4\sqrt{2}}\right) = \frac{3}{8}\left[1+\frac{1}{\sqrt{2}}\right],

    and the area of the rectangles on the right is:

    (12×122)+(12×14)+(12×142)+(12×18)=38[12+12].\left(\frac{1}{2}\times \frac{1}{2\sqrt{2}}\right)+ \left(\frac{1}{2}\times \frac{1}{4}\right)+ \left(\frac{1}{2}\times \frac{1}{4\sqrt{2}}\right)+ \left(\frac{1}{2}\times \frac{1}{8}\right) = \frac{3}{8}\left[\frac{1}{2}+\frac{1}{\sqrt{2}}\right].

    So, the area under the curve is in the interval (38[12+12],38[1+12])\left( \frac{3}{8}\left[\frac{1}{2}+\frac{1}{\sqrt{2}}\right], \frac{3}{8}\left[1+\frac{1}{\sqrt{2}}\right]\right). The length of this interval is 316\frac{3}{16}, and 316<315=15=0.2\frac{3}{16}<\frac{3}{15}=\frac{1}{5}=0.2, as desired. (Indeed, 316=0.1875<0.2\frac{3}{16}=0.1875<0.2.)

    Note, if we choose any value in the interval (38[12+12],38[1+12])\left( \frac{3}{8}\left[\frac{1}{2}+\frac{1}{\sqrt{2}}\right], \frac{3}{8}\left[1+\frac{1}{\sqrt{2}}\right]\right) as an approximation for the area under the curve, our error is no more than 0.2.

Q4Stage 1

Let f(x)f(x) be a function that is decreasing from x=0x=0 to x=5x=5. Which Riemann sum approximation of ∫05f(x)dx\displaystyle\int_0^5 f(x)\dee{x} is the largest–left, right, or midpoint?

Hint

Try drawing a picture.

Answer

left

Full solution

Since f(x)f(x) is decreasing, it is larger on the left endpoint of an interval than on the right endpoint of an interval. So, a left Riemann sum gives a larger approximation. Notice this does not depend on nn.

Furthermore, the actual area ∫05f(x)dx\displaystyle\int_0^5f(x)\dee{x} is larger than its right Riemann sum, and smaller than its left Riemann sum.

Figure from prob_s1.1, line 2

Figure from prob_s1.1, line 2

Figure from prob_s1.1, line 12

Figure from prob_s1.1, line 12

Q5Stage 1

Give an example of a function f(x)f(x), an interval [a,b][a,b], and a number nn such that the midpoint Riemann sum of f(x)f(x) over [a,b][a,b] using nn intervals is larger than both the left and right Riemann sums of f(x)f(x) over [a,b][a,b] using nn intervals.

Hint

Try an oscillating function.

Answer

Many answers are possible. One example is f(x)=sin⁡xf(x)=\sin x, [a,b]=[0,π][a,b]=[0,\pi], n=1n=1. Another example is f(x)=sin⁡xf(x)=\sin x, [a,b]=[0,5π][a,b]=[0,5\pi], n=5n=5.

Full solution

If f(x)f(x) is always increasing or always decreasing, then the midpoint Riemann sum will be between the left and right Riemann sums. So, we need a function that goes up and down. Many examples are possible, but let's work with a familiar one: sin⁡x\sin x.

If our intervals have endpoints that are integer multiples of π\pi, then the left and right Riemann sums will be 0, since sin⁡(0)=sin⁡(π)=sin⁡(2π)=⋯=0\sin(0)=\sin(\pi)=\sin(2\pi)=\cdots=0. The midpoints of these intervals will give yy-values of 1 and -1. So, for example, we can let f(x)=sin⁡xf(x)=\sin x, [a,b]=[0,π][a,b]=[0,\pi], and n=1n=1. Then the right and left Riemann sums are 0, while the midpoint Riemann sum is π\pi.

We can extend the example of f(x)=sin⁡xf(x)=\sin x to have more intervals. As long as we have more positive terms than negative, the midpoint approximation will be a positive number, and so it will be larger than both the left and right Riemann sums. So, for example, we can let f(x)=sin⁡xf(x)=\sin x, [a,b]=[0,5π][a,b]=[0,5\pi], and n=5n=5. Then the midpoint Riemann sum is π−π+π−π+π=π\pi-\pi+\pi-\pi+\pi=\pi, which is strictly larger than 0 and so it is larger than both the left and right Riemann sums.

Figure from prob_s1.1, line 2

Figure from prob_s1.1, line 2

In Questions 6 through 10, we practice using sigma notation. There are many ways to write a given sum in sigma notation. You can practice finding several, and deciding which looks the clearest.

Q6Stage 1

Express the following sums in sigma notation:

  1. 3+4+5+6+73+4+5+6+7

  2. 6+8+10+12+146+8+10+12+14

  3. 7+9+11+13+157+9+11+13+15

  4. 1+3+5+7+9+11+13+151+3+5+7+9+11+13+15

Hint

The ordering of the parts is intentional: each sum can be written by changing some small part of the sum before it.

Answer

Some of the possible answers are given, but more exist.

  1. ∑i=37i\displaystyle\sum_{i=3}^7 i ; ∑i=15(i+2)\displaystyle\sum_{i=1}^5 (i+2)

  2. ∑i=372i\displaystyle\sum_{i=3}^7 2i ; ∑i=15(2i+4)\displaystyle\sum_{i=1}^5 (2i+4)

  3. ∑i=37(2i+1)\displaystyle\sum_{i=3}^7 (2i+1) ; ∑i=15(2i+5)\displaystyle\sum_{i=1}^5 (2i+5)

  4. ∑i=18(2i−1)\displaystyle\sum_{i=1}^8 (2i-1) ; ∑i=07(2i+1)\displaystyle\sum_{i=0}^7 (2i+1)

Full solution

  1. Two possible answers are ∑i=37i\displaystyle\sum_{i=3}^7 i and ∑i=15(i+2)\displaystyle\sum_{i=1}^5 (i+2). The first has simpler terms (ii versus i+2i+2), while the second has simpler indices (we often like to start at i=1i=1). Neither is objectively better than the other, but depending on your purposes you might find one more useful.

  2. The terms of this sum are each double the terms of the sum from part (a), so two possible answers are ∑i=372i\displaystyle\sum_{i=3}^7 2i and ∑i=15(2i+4)\displaystyle\sum_{i=1}^5 (2i+4).
    We often want to write a sum that involves even numbers: it will be useful for you to remember that the term 2i2i (with index ii) generates evens.

  3. The terms of this sum are each one more than the terms of the sum from part (b), so two possible answers are ∑i=37(2i+1)\displaystyle\sum_{i=3}^7 (2i+1) and ∑i=15(2i+5)\displaystyle\sum_{i=1}^5 (2i+5).
    In the last part, we used the expression 2i2i to generate even numbers; 2i+12i+1 will generate odds. So will the index 2i+52i+5, and indeed, 2i+k2i+k for any odd number kk. The choice of what you add will depend on the limits of ii.

  4. This sum adds up the odd numbers from 1 to 15. From Part (c), we know that the formula 2i+12i+1 is a simple way of generating odd numbers. Since our first term should be 1 and our last term should be 15, if we use ∑(2i+1)\sum (2i+1), then ii should run from 00 to 77. So, one way of expressing our sum in sigma notation is ∑i=07(2i+1)\displaystyle\sum_{i=0}^7 (2i+1).

    Sometimes we like our sum to start at i=1i=1 instead of i=0i=0. If this is our desire, we can use 2i−12i-1 as our terms, and let ii run from 1 to 8. This gives us another way of expressing our sum: ∑i=18(2i−1)\displaystyle\sum_{i=1}^8 (2i-1).

Q7Stage 1

Express the following sums in sigma notation:

  1. 13+19+127+181\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}

  2. 23+29+227+281\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+\frac{2}{81}

  3. −23+29−227+281-\frac{2}{3}+\frac{2}{9}-\frac{2}{27}+\frac{2}{81}

  4. 23−29+227−281\frac{2}{3}-\frac{2}{9}+\frac{2}{27}-\frac{2}{81}

Hint

If we raise −1-1 to an even power, we get +1+1, and if we raise it to an odd power, we get −1-1.

Answer

Some answers are below, but others are possible.

  1. ∑i=1413i\displaystyle\sum_{i=1}^4 \frac{1}{3^i} ; ∑i=14(13)i\displaystyle\sum_{i=1}^4 \left(\frac{1}{3}\right)^i

  2. ∑i=1423i\displaystyle\sum_{i=1}^4 \frac{2}{3^i} ; ∑i=142(13)i\displaystyle\sum_{i=1}^4 2\left(\frac{1}{3}\right)^i

  3. ∑i=14(−1)i23i\displaystyle\sum_{i=1}^4(-1)^i \frac{2}{3^i} ; ∑i=142(−3)i\displaystyle\sum_{i=1}^4 \frac{2}{(-3)^i}

  4. ∑i=14(−1)i+123i\displaystyle\sum_{i=1}^4(-1)^{i+1} \frac{2}{3^i} ; ∑i=14−2(−3)i\displaystyle\sum_{i=1}^4 -\frac{2}{(-3)^i}

Full solution

  1. The denominators are successive powers of three, so one way of writing this is ∑i=1413i\displaystyle\sum_{i=1}^4 \frac{1}{3^i}. Equivalently, the terms we're adding are powers of 1/31/3, so we can also write ∑i=14(13)i\displaystyle\sum_{i=1}^4 \left(\frac{1}{3}\right)^i.

  2. This sum is obtained from the sum in (a) by multiplying each term by two, so we can write ∑i=1423i\displaystyle\sum_{i=1}^4 \frac{2}{3^i} or ∑i=142(13)i\displaystyle\sum_{i=1}^4 2\left(\frac{1}{3}\right)^i.

  3. The difference between this sum and the previous sum is its alternating sign, minus-plus-minus-plus. This behaviour appears when we raise a negative number to successive powers. We can multiply each term by (−1)i(-1)^i, or we can slip a negative into the number that is already raised to the power ii: ∑i=14(−1)i23i\displaystyle\sum_{i=1}^4(-1)^i \frac{2}{3^i} , or ∑i=142(−3)i\displaystyle\sum_{i=1}^4 \frac{2}{(-3)^i}.

  4. This sum is the negative of the sum in part (c), so we can simply multiply each term by negative one: ∑i=14(−1)i+123i\displaystyle\sum_{i=1}^4(-1)^{i+1} \frac{2}{3^i} , or ∑i=14−2(−3)i\displaystyle\sum_{i=1}^4 -\frac{2}{(-3)^i} .

    Be careful with the second form: a common mistake is to think that −2(−3)i=23i-\dfrac{2}{(-3)^i} = \dfrac{2}{3^i}, but these are not the same.

Q8Stage 1

Express the following sums in sigma notation:

  1. 13+13+527+781+9243\frac{1}{3}+\frac{1}{3}+\frac{5}{27}+\frac{7}{81}+\frac{9}{243}

  2. 15+111+129+183+1245\frac{1}{5}+\frac{1}{11}+\frac{1}{29}+\frac{1}{83}+\frac{1}{245}

  3. 1000+200+30+4+12+350+710001000+200+30+4+\frac{1}{2}+\frac{3}{50}+\frac{7}{1000}

Hint

Sometimes a little anti-simplification can make the pattern more clear.

  1. Re-write as 13+39+527+781+9243\frac{1}{3}+\frac{3}{9}+\frac{5}{27}+\frac{7}{81}+\frac{9}{243}.

  2. Compare to the sum in the hint for (a).

  3. Re-write as 1⋅1000+2⋅100+3⋅10+41+510+6100+710001\cdot1000+2\cdot 100+3\cdot10+\frac{4}{1}+\frac{5}{10}+\frac{6}{100}+\frac{7}{1000}.

Answer

  1. ∑i=152i−13i\displaystyle\sum_{i=1}^5 \frac{2i-1}{3^i}

  2. ∑i=1513i+2\displaystyle\sum_{i=1}^5 \frac{1}{3^i+2}

  3. ∑i=17i⋅104−i\displaystyle\sum_{i=1}^7 i\cdot10^{4-i} ; ∑i=17i10i−4\displaystyle\sum_{i=1}^7 \frac{i}{10^{i-4}}

Full solution

  1. If we re-write the second term as 39\frac{3}{9} instead of 13\frac{1}{3}, our sum becomes:

    13+39+527+781+9243\frac{1}{3}+\frac{3}{9}+\frac{5}{27}+\frac{7}{81}+\frac{9}{243}

    The numerators are the first five odd numbers, and the denominators are the first five positive powers of 3. We learned how to generate odd numbers in Question 6, and we learned how to generate powers of three in Question 7. Combining these, we can write our sum as ∑i=152i−13i\displaystyle\sum_{i=1}^5 \frac{2i-1}{3^i} .

  2. The denominators of these terms differ from the denominators of part (a) by precisely two, while the numerators are simply 1. So, we can modify our previous answer: ∑i=1513i+2\displaystyle\sum_{i=1}^5 \frac{1}{3^i+2} .

  3. Let's re-write the sum to make the pattern clearer.

1000+200+30+4+12+350+71000=1⋅1000+2⋅100+3⋅10+41+510+6100+71000=1⋅103+2⋅102+3⋅101+4⋅100+5⋅10−1+6⋅10−2+7⋅10−3=1⋅104−1+2⋅104−2+3⋅104−3+4⋅104−4+5⋅104−5+6⋅104−6+7⋅104−7\begin{array}{cccccccccccccc} &1000&+&200&+&30&+&4&+&\frac{1}{2}&+&\frac{3}{50}&+&\frac{7}{1000}\\[5pt] =&1\cdot1000&+&2\cdot 100&+&3\cdot10&+&\frac{4}{1}&+&\frac{5}{10}&+&\frac{6}{100}&+&\frac{7}{1000}\\[5pt] =&1\cdot 10^3 &+& 2\cdot 10^2&+&3\cdot 10^1 &+& 4\cdot 10^0 &+& 5 \cdot 10^{-1} &+& 6\cdot 10^{-2} &+& 7 \cdot 10^{-3} \\[5pt] =&\textcolor{red}1\cdot 10^{4-\textcolor{red}1} &+& \textcolor{red}2\cdot 10^{4-\textcolor{red}2}&+&\textcolor{red}3\cdot 10^{4-\textcolor{red}3} &+& \textcolor{red}4\cdot 10^{4-\textcolor{red}4} &+& \textcolor{red}5 \cdot 10^{4-\textcolor{red}5} &+& \textcolor{red}6\cdot 10^{4-\textcolor{red}6} &+& \textcolor{red}7 \cdot 10^{4-\textcolor{red}7} \end{array}

If we let the red numbers be our index ii, this gives us the expression ∑i=17i⋅104−i\displaystyle\sum_{i=1}^7 i\cdot10^{4-i} . Equivalently, we can write ∑i=17i10i−4\displaystyle\sum_{i=1}^7 \frac{i}{10^{i-4}} .

Q9Stage 1

Evaluate the following sums. You might want to use the formulas from Theorems  1.1.5 and 1.1.6 in the CLP-2 text.

  1. ∑i=0100(35)i\displaystyle\sum_{i=0}^{100} \left(\dfrac{3}{5}\right)^i

  2. ∑i=50100(35)i\displaystyle\sum_{i=50}^{100} \left(\dfrac{3}{5}\right)^i

  3. ∑i=110(i2−3i+5)\displaystyle\sum_{i=1}^{10} \left(i^2-3i+5\right)

  4. ∑n=1b[(1e)n+en3]\displaystyle\sum_{n=1}^{b}\left[ \left(\frac{1}{e}\right)^n+en^3\right], where bb is some integer greater than 1.

Hint

(a), (b) These are geometric sums.
(c) You can write this as three separate sums.
(d) You can write this as two separate sums. Remember that ee is a constant. Don't be thrown off by the index being nn instead of ii.

Answer

  1. 52[1−(35)101]\dfrac{5}{2}\left[1-\left(\dfrac{3}{5}\right)^{101}\right]

  2. 52(35)50[1−(35)51]\dfrac{5}{2}\left(\dfrac{3}{5}\right)^{50}\left[1-\left(\dfrac{3}{5}\right)^{51}\right]

  3. 270270

  4. 1−(1e)be−1+e4[b(b+1)]2\dfrac{1-\left(\frac{1}{e}\right)^b}{e-1}+\dfrac{e}{4}\left[b(b+1)\right]^2

Full solution

  1. Using Theorem 1.1.6.a in the CLP-2 text, with a=1a=1, r=35r=\frac{3}{5} and n=100n=100:

    ∑i=0100(35)i=1−(35)1011−35=52[1−(35)101]\sum_{i=0}^{100} \left(\dfrac{3}{5}\right)^i = \dfrac{1-\left(\frac{3}{5}\right)^{101}}{1-\frac{3}{5}} = \dfrac{5}{2}\left[1-\left(\frac{3}{5}\right)^{101}\right]

  2. We want to use Theorem 1.1.6, part (a) again, but our sum doesn't start at (35)0=1\left(\frac{3}{5}\right)^0=1. We have two options: factor out the leading term, or use the difference of two sums that start where we want them to.

    • In this solution, we'll make our sum start at 1 by factoring out the leading term. We wrote our work out the long way (expanding the sigma into “dot-dot-dot" notation) for clarity, but it's faster to do the algebra in sigma notation all the way through.

      ∑i=50100(35)i=(35)50+(35)51+(35)52+⋯+(35)100=(35)50[1+(35)+(35)2+⋯+(35)50]=(35)501−(35)511−35=52(35)50[1−(35)51].\begin{align*} \displaystyle\sum_{i=50}^{100} \left(\dfrac{3}{5}\right)^i&= \left(\dfrac{3}{5}\right)^{50}+ \left(\dfrac{3}{5}\right)^{51}+ \left(\dfrac{3}{5}\right)^{52}+\cdots+ \left(\dfrac{3}{5}\right)^{100}\\ &= \left(\dfrac{3}{5}\right)^{50}\left[1+ \left(\dfrac{3}{5}\right)+ \left(\dfrac{3}{5}\right)^{2}+\cdots+ \left(\dfrac{3}{5}\right)^{50}\right] \\ &= \left(\dfrac{3}{5}\right)^{50}\dfrac{1-\left(\frac{3}{5}\right)^{51}}{1-\frac{3}{5}}\\ &=\dfrac{5}{2}\left(\dfrac{3}{5}\right)^{50}\left[1-\left(\frac{3}{5}\right)^{51}\right]. \end{align*}

    • In this solution, we write our given expression as the difference of two sums, both starting at i=0i=0.

      ∑i=50100(35)i=∑i=0100(35)i−∑i=049(35)i=1−(35)1011−35−1−(35)501−35=52[(35)50−(35)101]=52(35)50[1−(35)51].\begin{align*} \displaystyle\sum_{i=50}^{100} \left(\dfrac{3}{5}\right)^i&= \displaystyle\sum_{i=0}^{100} \left(\dfrac{3}{5}\right)^i- \displaystyle\sum_{i=0}^{49} \left(\dfrac{3}{5}\right)^i\\ &=\dfrac{1-\left(\frac{3}{5}\right)^{101}}{1-\frac{3}{5}} - \dfrac{1-\left(\frac{3}{5}\right)^{50}}{1-\frac{3}{5}} \\ &=\dfrac{5}{2}\left[\left(\frac{3}{5}\right)^{50}-\left(\frac{3}{5}\right)^{101}\right]\\ &=\dfrac{5}{2}\left(\dfrac{3}{5}\right)^{50}\left[1-\left(\frac{3}{5}\right)^{51}\right]. \end{align*}

  3. Before we can use the equations in Theorem 1.1.6, we'll need to do a little simplification.

    ∑i=110(i2−3i+5)=∑i=110i2+∑i=110−3i+∑i=1105=∑i=110i2−3∑i=110i+5∑i=1101=16(10)(11)(21)−3(12(10⋅11))+5⋅10=270\begin{align*} \displaystyle\sum_{i=1}^{10} \left(i^2-3i+5\right)&= \displaystyle\sum_{i=1}^{10} i^2 +\displaystyle\sum_{i=1}^{10} -3i +\displaystyle\sum_{i=1}^{10}5\\ &= \displaystyle\sum_{i=1}^{10} i^2 -3\displaystyle\sum_{i=1}^{10} i +5\displaystyle\sum_{i=1}^{10}1\\ &= \frac{1}{6}(10)(11)(21) -3\left(\frac{1}{2}(10\cdot 11)\right) +5\cdot 10\\ &=270 \end{align*}

  4. As in part (c), we'll simplify first. The first part (shown here in red) is a geometric sum, but it does not start at 1=(1e)01=\left(\frac{1}{e}\right)^0.

    ∑n=1b[(1e)n+ en3]=∑n=1b(1e)n+∑n=1ben3=∑n=0b(1e)n−1+e∑n=1bn3=1−(1e)b+11−1e−1+e[12b(b+1)]2=1e−(1e)b+11−1e+e[12b(b+1)]2=1−(1e)be−1+e4[b(b+1)]2\begin{align*} \displaystyle\sum_{n=1}^{b}\left[\textcolor{red}{ \left(\frac{1}{e}\right)^n}\textcolor{black}+ \,\textcolor{blue}{en^3}\right]&= \color{red}\displaystyle\sum_{n=1}^{b} \left(\frac{1}{e}\right)^n\color{black}+\color{blue} \displaystyle\sum_{n=1}^{b}en^3\\ &=\color{red} \displaystyle\sum_{n=0}^{b} \left(\frac{1}{e}\right)^{n}-1 \color{black}+ \color{blue}e\displaystyle\sum_{n=1}^{b}n^3\\ &=\color{red} \dfrac{1-\left(\frac{1}{e}\right)^{b+1}}{1-\frac{1}{e}}-1 \color{black}+ \color{blue}e\left[\frac{1}{2}b(b+1)\right]^2\\ &=\color{red} \dfrac{\frac{1}{e}-\left(\frac{1}{e}\right)^{b+1}}{1-\frac{1}{e}} \color{black}+ \color{blue}e\left[\frac{1}{2}b(b+1)\right]^2\\ &=\textcolor{red}{\dfrac{1-\left(\frac{1}{e}\right)^b}{e-1}}+\textcolor{blue}{\frac{e}{4}\left[b(b+1)\right]^2} \end{align*}
Q10Stage 1

Evaluate the following sums. You might want to use the formulas from Theorem 1.1.6 in the CLP-2 text.

  1. ∑i=50100(i−50)+∑i=050i\displaystyle\sum_{i=50}^{100} (i-50)+\displaystyle\sum_{i=0}^{50} i

  2. ∑i=10100(i−5)3\displaystyle\sum_{i=10}^{100} \left(i-5\right)^3

  3. ∑n=111(−1)n\displaystyle\sum_{n=1}^{11} (-1)^n

  4. ∑n=211(−1)2n+1\displaystyle\sum_{n=2}^{11} (-1)^{2n+1}

Hint

  1. Write out the terms of the two sums.

  2. A change of index is an easier option than expanding the cubic.

  3. Which terms cancel?

  4. Remember 2n+12n+1 is odd for every integer nn. The index starts at n=2n=2, not n=1n=1.

Answer

  1. 50⋅51=255050\cdot 51=2550

  2. [12(95)(96)]2−[12(4)(5)]2=20,793,500\left[\frac{1}{2}(95)(96)\right]^2-\left[\frac{1}{2}(4)(5)\right]^2=20,793,500

  3. −1-1

  4. −10-10

Full solution

  1. The two pieces are very similar, which we can see by changing the index, or expanding them out:

    ∑i=50100(i−50)+∑i=050i=(0+1+2+⋯+50)+(0+1+2+⋯+50)=(1+2+⋯+50)+(1+2+⋯+50)=2(1+2+⋯+50)=2∑i=150i=2(50⋅512)=50⋅51=2550\begin{align*} \displaystyle\sum_{i=50}^{100} (i-50)+\displaystyle\sum_{i=0}^{50} i&= \left(0+1+2+\cdots + 50\right)+\left(0+1+2+\cdots + 50\right)\\ &=\left(1+2+\cdots + 50\right)+\left(1+2+\cdots + 50\right)\\ &=2\left(1+2+\cdots + 50\right)\\ &=2\sum_{i=1}^{50} i\\ &= 2\left(\frac{50\cdot 51}{2}\right)=50\cdot 51=2550 \end{align*}

  2. If we expand (i−5)3=i3−15i2+75i−125(i-5)^3 = i^3-15i^2+75i-125, we can break the sum into four parts, and evaluate each separately. However, it is much simpler to change the index and make the term (i−5)3(i-5)^3 into i3i^3.

    ∑i=10100(i−5)3=53+63+73+⋯+953\begin{align*}\displaystyle\sum_{i=10}^{100} \left(i-5\right)^3&= 5^3+6^3+7^3+\cdots +95^3\end{align*}

    We have a formula to evaluate the sum of cubes if they start at 11, so we turn our expression into the difference of two sums starting at 1:

    =[13+23+33+43+53+63+73+⋯+953]−[13+23+33+43]=∑i=195i3−∑i=14i3=[12(95)(96)]2−[12(4)(5)]2=20,793,500 .\begin{align*}&= \left[1^3+2^3+3^3+4^3+5^3+6^3+7^3+\cdots +95^3\right]- \left[1^3+2^3+3^3+4^3\right]\\ &=\displaystyle\sum_{i=1}^{95} i^3 - \displaystyle\sum_{i=1}^4 i^3\\ &=\left[\frac{1}{2}(95)(96)\right]^2-\left[\frac{1}{2}(4)(5)\right]^2\\ &=20,793,500\,.\end{align*}

  3. Notice every two terms cancel with each other, since the sum is (−1)+(+1)(-1)+(+1), etc. Then the terms n=1n=1 through n=10n=10 cancel, and we're left only with the final term, (−1)11=−1(-1)^{11}=-1.

    Written out more explicitly:

    ∑n=111(−1)n=−1+1−1+1−1+1−1+1−1+1−1=[−1+1]+[−1+1]+[−1+1]+[−1+1]+[−1+1]−1=0+0+0+0+0−1=−1.\begin{align*} \displaystyle\sum_{n=1}^{11} (-1)^{n}&=-1+1-1+1-1+1-1+1-1+1-1\\ &=[-1+1]+[-1+1]+[-1+1]+[-1+1]+[-1+1]-1\\ &=0+0+0+0+0-1=-1. \end{align*}

  4. For every integer nn, 2n+12n+1 is odd, so (−1)2n+1=−1(-1)^{2n+1}=-1. Then ∑n=211(−1)2n+1=∑n=211−1=−10\displaystyle\sum_{n=2}^{11} (-1)^{2n+1} =\displaystyle\sum_{n=2}^{11} -1 =-10.

Questions 11 through 15 are meant to give you practice interpreting the formulas in Definition 1.1.11 of the CLP-2 text. The formulas might look complicated at first, but if you understand what each piece means, they are easy to learn.

Q11Stage 1

In the picture below, draw in the rectangles whose (signed) area is being computed by the midpoint Riemann sum ∑i=14b−a4⋅f(a+(i−12)b−a4)\displaystyle\sum_{i=1}^4 \dfrac{b-a}{4}\cdot f\left(a+\left(i-\frac{1}{2}\right)\dfrac{b-a}{4}\right).

Figure from prob_s1.1, line 2

Figure from prob_s1.1, line 2

Hint

Since the sum adds four pieces, there will be four rectangles. However, one might be extremely small.

Answer

Figure from prob_s1.1, line 2

Figure from prob_s1.1, line 2

Full solution

The index of the sum runs from 1 to 4: the first, second, third, and fourth rectangles. So, we have four rectangles in our Riemann sum. Let's start by drawing in the intervals along the xx-axis taken up by these four rectangles. Note each has the same width: b−a4\dfrac{b-a}{4}.

Figure from prob_s1.1, line 2

Figure from prob_s1.1, line 2

Since this is a midpoint Riemann sum, the height of each rectangle is given by the yy-value of the function in the midpoint of the interval. So, now let's find the height of the function at the midpoints of each of the four intervals.

Figure from prob_s1.1, line 2

Figure from prob_s1.1, line 2

The left-most interval has a height of about 0, so it gives a “trivial" rectangle with no height and no area. The middle two intervals have rectangles of about the same height, and the right-most interval has the highest rectangle.

Figure from prob_s1.1, line 2

Figure from prob_s1.1, line 2

Q12Stage 1Past exam · M105 2015A

∑k=14f(1+k)⋅1\displaystyle \sum_{k=1}^4 f(1+k)\cdot 1 is a left Riemann sum for a function f(x)f(x) on the interval [a,b][a,b] with nn subintervals. Find the values of aa, bb and nn.

Hint

Write out the general formula for the left Riemann sum from Definition 1.1.11 in the CLP-2 text and choose aa, bb and nn to make it match the given sum.

Answer

n=4n=4, a=2a=2, and b=6b=6

Full solution

In general, the left Riemann sum for the integral ∫abf(x)  dx\int_a^b f(x)\,\,\dee{x} is of the form

∑k=1nf(a+(k−1)b−an)b−an\begin{align*} \sum_{k=1}^n f\left(a+(k-1)\frac{b-a}{n}\right)\frac{b-a}{n} \end{align*}

  • To get the limits of summation to match the given sum, we need n=4n=4.

  • Then to get the factor multiplying ff to match that in the given sum, we need b−an=1\frac{b-a}{n}=1, so b−a=4b-a=4.

  • Finally, to get the argument of ff to match that in the given sum, we need

    a+(k−1)b−an=a−b−an+kb−an=1+k\begin{align*} a+(k-1)\frac{b-a}{n}=a-\frac{b-a}{n} +k\frac{b-a}{n}=1+k \end{align*}

    Subbing in n=4n=4 and b−a=4b-a=4 gives a−1+k=1+ka-1 +k=1+k, so a=2a=2 and b=6b=6.

Q13Stage 1

Draw a picture illustrating the area given by the following Riemann sum.

∑i=132⋅(5+2i)2\sum_{i=1}^3 2\cdot\left(5+2i\right)^2
Hint

Since the sum runs from 1 to 3, there are three intervals. Suppose 2=Δx=b−an2 = \Delta x = \frac{b-a}{n}. You may assume the sum given is a right Riemann sum (as opposed to left or midpoint).

Answer

One answer is below, but other interpretations exist.

Figure from prob_s1.1, line 2

Figure from prob_s1.1, line 2

Full solution

The general form of a Riemann sum is ∑i=1nΔx⋅f(xi∗)\displaystyle\sum_{i=1}^n \Delta x \cdot f(x_i^*), where Δx=b−an\Delta x = \frac{b-a}{n} is the width of each rectangle, and f(xi∗)f(x_i^*) is the height.

There are different ways to interpret the given sum as a Riemann sum. The most obvious is given in Solution 1. You may notice that we make some convenient assumptions in this solution about values for Δx\Delta x and aa, and we assume the sum is a right Riemann sum. Other visualizations of the sum arise from making more exotic choices. Some of these are explored in Solutions 2-4.

All cases have three rectangles, and the three rectangles will have the same areas: 98, 162, and 242 square units, respectively. This is because the terms of the given sum simplify to 98+162+24298+162+242.

    • Because the index runs from 11 to 33, there are three intervals: n=3n=3.

    • Looking at our sum, it seems reasonable to interpret Δx=2\Delta x = 2. Then, since n=3n=3, we conclude b−a3=2\frac{b-a}{3}=2, hence b−a=6b-a=6.

    • If Δx=2\Delta x = 2, then f(xi∗)=(5+2i)2f(x_i^*)=\left(5+2i\right)^2. Recall that xi∗x_i^* is the xx-coordinate we use to decide the height of the iith rectangle. In a right Riemann sum, xi∗=a+i⋅Δxx_i^* = a+i\cdot\Delta x. So, using 2=Δx2=\Delta x, we can let f(xi∗)=f(a+2i)=(5+2i)2f(x_i^*)=f(a+2i)=\left(5+2i\right)^2. This fits with the function f(x)=x2f(x)=x^2, and a=5a=5.

    • Since b−a=6b-a=6, and a=5a=5, this tells us b=11b=11

    To sum up, we can interpret the Riemann sum as a right Riemann sum, with three intervals, of the function f(x)=x2f(x)=x^2 from x=5x=5 to x=11x=11.

    Figure from prob_s1.1, line 2

    Figure from prob_s1.1, line 2

    • The index of the sum runs from 1 to 3, so we have n=3n=3.

    • We didn't have to interpret Δx\Delta x as 2–that was just the path of least resistance. We could have chosen it to be any other number–for the sake of argument, let's say Δx=10\Delta x=10. (Positive numbers are easiest to interpret, but negatives are technically allowed as well.)

    • Then 10=b−an=b−a310=\frac{b-a}{n}=\frac{b-a}{3}, so b−a=30b-a=30.

    • Let's use the paradigm of a right Riemann sum, and match up the terms of the sum given in the problem to the terms in the definition:

      Δx⋅f(a+i⋅Δx)=2⋅(5+2i)210⋅f(a+10i)=2⋅(5+2i)2f(a+10i)=15⋅(5+2i)2f(a+10i)=15⋅(5+15⋅10i)2\begin{align*} \Delta x \cdot f\left(a+i\cdot \Delta x\right)&= 2\cdot\left(5+2i\right)^2\\ 10 \cdot f(a+10i)&= 2\cdot\left(5+2i\right)^2\\ f(a+10i)&=\frac{1}{5}\cdot\left(5+2i\right)^2\\ f(a+10i)&=\frac{1}{5}\cdot\left(5+\frac{1}{5}\cdot 10i\right)^2 \end{align*}

    • The easiest value of aa in this case is a=0a=0. Then f(10i)=15⋅(5+15⋅10i)2f(\textcolor{red}{10i}) = \frac{1}{5}\cdot\left(5+\frac{1}{5}\cdot \textcolor{red}{10i}\right)^2, so f(x)=15⋅(5+15⋅x)2f(\textcolor{red}{x})= \frac{1}{5}\cdot\left(5+\frac{1}{5}\cdot \textcolor{red}{x}\right)^2.

    • If a=0a=0 and b−a=30b-a=30, then b=30b=30.

    • To sum up: n=3n=3, a=0a=0, b=30b=30, Δx=10\Delta x = 10, and f(x)=15⋅(5+x5)2f(x)= \frac{1}{5}\cdot\left(5+\frac{x}{5} \right)^2.

      Figure from prob_s1.1, line 2

      Figure from prob_s1.1, line 2

      By changing Δx\Delta x, we changed the widths of the rectangles. The rectangles in this picture are wider and shorter than the rectangles in Solution 1. Their areas are the same: 98, 162, and 242.

    • Suppose Δx=2\Delta x = 2, and we interpret our sum as a right Riemann sum, but we didn't assume a=5a=5. We could have chosen aa to be any number–say, a=1a=1.

    • Let's match up what we're given in the problem to what we're given as a definition:

      Δx⋅f(a+i⋅Δx)=2⋅(5+2i)22⋅f(1+2i)=2⋅(5+2i)2f(1+2i)=(5+2i)2f(1+2i)=(4+1+2i)2\begin{align*} \Delta x \cdot f\left(a+i\cdot\Delta x\right)&=2\cdot\left(5+2i\right)^2\\ 2 \cdot f\left(1+2i\right)&=2\cdot\left(5+2i\right)^2\\ f\left(1+2i\right)&=\left(5+2i\right)^2\\ f\left(1+2i\right)&=\left(4+1+2i\right)^2 \end{align*}

    • Since f(1+2i)=(4+1+2i)2f(\textcolor{red}{1+2i})=\left(4+\textcolor{red}{1+2i}\right)^2, we have f(x)=(4+x)2f(\textcolor{red}{x})=\left(4+\textcolor{red}{x}\right)^2

    • Since a=1a=1 and b−a3=2\frac{b-a}{3}=2, in this case b=7b=7.

    • To sum up: n=3n=3, a=1a=1, b=7b=7, Δx=2\Delta x=2, and f(x)=(4+x)2f(x)=(4+x)^2.

      Figure from prob_s1.1, line 2

      Figure from prob_s1.1, line 2

      This picture is a lot like the picture in Solution 1, but shifted to the left. By changing aa, we changed the left endpoint of our region.

    • We didn't have to assume that we were dealing with a right Riemann sum. Suppose Δx=2\Delta x =2, and we have a midpoint Riemann sum.

    • Let's match up what we're given in the problem with what we're given in the definition:

      Δx⋅f(a+(i−12)Δx)=2⋅(5+2i)22⋅f(a+(i−12)2)=2⋅(5+2i)2f(a+(i−12)2)=(5+2i)2f(a+2i−1)=(5+2i)2f((a−1)+2i)=(5+2i)2\begin{align*} \Delta x \cdot f\left(a+\left(i-\tfrac{1}{2}\right)\Delta x\right)&=2\cdot\left(5+2i\right)^2\\ 2 \cdot f\left(a+\left(i-\tfrac{1}{2}\right)2\right)&=2\cdot\left(5+2i\right)^2\\ f\left(a+\left(i-\tfrac{1}{2}\right)2\right)&=\left(5+2i\right)^2\\ f\left(a+2i-1\right)&=\left(5+2i\right)^2\\ f\left((a-1)+2i\right)&=\left(5+2i\right)^2 \end{align*}

    • It is now convenient to set a−1=5a-1=5, hence a=6a=6.

    • Then f(5+2i)=(5+2i)2f(\textcolor{red}{5+2i})=(\textcolor{red}{5+2i})^2, so f(x)=x2f(\textcolor{red}{x})=\textcolor{red}{x}^2

    • Since 2=b−a32=\frac{b-a}{3} and a=6a=6, we see b=12b=12.

    • To sum up: n=3n=3, a=6a=6, b=12b=12, Δx=2\Delta x = 2, and f(x)=x2f(x)=x^2.

      Figure from prob_s1.1, line 2

      Figure from prob_s1.1, line 2

      By choosing to interpret our sum as a midpoint Riemann sum instead of a right Riemann sum, we changed where our rectangles intersect the graph y=f(x)y=f(x): instead of the graph hitting the right corner of the rectangle, it hits in the middle.

Q14Stage 1

Draw a picture illustrating the area given by the following Riemann sum.

∑i=15π20⋅tan⁡(π(i−1)20)\sum_{i=1}^5 \frac{\pi}{20}\cdot \tan\left(\frac{\pi (i-1)}{20}\right)
Hint

Let Δx=π20\Delta x = \dfrac{\pi}{20}. Then what is b−ab-a?

Answer

Many interpretations are possible–see the solution to Question 13 for a more thorough discussion–but the most obvious is given below.

Figure from prob_s1.1, line 2

Figure from prob_s1.1, line 2

Full solution

Many interpretations are possible–see the solution to Question 13 for a more thorough discussion–but the most obvious is given below. Recall the definition of a left Riemann sum:

∑i=1nΔx⋅f(a+(i−1)Δx)\sum_{i=1}^n \Delta x \cdot f\left(a+(i-1)\Delta x\right)

We chose a left Riemann sum instead of right or midpoint because our given sum has (i−1)(i-1) in it, rather than (i−12)(i-\frac{1}{2}) or simply ii.

  • Since the sum has five terms (ii runs from 1 to 5), there are 5 rectangles. That is, n=5n=5.

  • In the definition of the Riemann sum, note that the term Δx\Delta x appears twice: once multiplied by the entire term, and once multiplied by i−1i-1. So, a convenient choice for Δx\Delta x is π20\frac{\pi}{20}, because this is the constant that is both multiplied at the start of the term, and multiplied by i−1i-1.

  • Since π20=Δx=b−an=b−a5\dfrac{\pi}{20}=\Delta x = \dfrac{b-a}{n} = \dfrac{b-a}{5}, we see b−a=5π20=π4b-a=\dfrac{5\pi}{20}=\dfrac{\pi}{4}.

  • We match the terms in the definition with the terms in the problem:

    f(a+(i−1)Δx)=tan⁡(π(i−1)20)f(a+(i−1)π20)=tan⁡((i−1)π20)\begin{align*} f(a+(i-1)\Delta x) & = \tan\left(\frac{\pi (i-1)}{20}\right)\\ f\left(a+(i-1)\frac{\pi}{20}\right) & = \tan\left((i-1)\frac{\pi }{20}\right) \end{align*}

    So, we choose a=0a=0 and f(x)=tan⁡xf(x) = \tan x.

  • Since a=0a=0 and b−a=π4b-a=\frac{\pi}{4}, we see b=π4b=\frac{\pi}{4}.

Figure from prob_s1.1, line 2

Figure from prob_s1.1, line 2

We note that the first rectangle of the five is a “trivial" rectangle, with height (and area) 0.

Q15Stage 1Past exam · M105 2013A

Fill in the blanks with right, left, or midpoint; an interval; and a value of n.

  1. ∑k=03f(1.5+k)⋅1\sum\limits_{k=0}^3 f (1.5 + k) \cdot 1 is a             ‾\underline{\ \ \ \ \ \ \ \ \ \ \ \ } Riemann sum for ff on the interval [       ‾ ,       ‾ ][\,\underline{\ \ \ \ \ \ }\ ,\ \underline{\ \ \ \ \ \ }\,] with n=     ‾n =\underline{\ \ \ \ \ }.

Hint

Notice that the index starts at k=0k=0, instead of k=1k=1. Write out the given sum explicitly without using summation notation, and sketch where the rectangles would fall on a graph of y=f(x)y=f(x).

Then try to identify b−ab-a, and nn, followed by “right”, “left”, or “midpoint”, and finally aa.

Answer

Three answers are possible. It is a midpoint Riemann sum for ff on the interval [1,5][1,5] with n=4n =4. It is also a left Riemann sum for ff on the interval [1.5,5.5][1.5,5.5] with n=4n =4. It is also a right Riemann sum for ff on the interval [0.5,4.5][0.5,4.5] with n=4n =4.

Full solution

Since there are four terms in the sum, n=4n=4. (Note the sum starts at k=0k=0, instead of k=1k=1.) Since the function is multiplied by 1, 1=Δx=b−an=b−a41=\Delta x=\dfrac{b-a}{n}=\dfrac{b-a}{4}, hence b−a=4b-a=4.

We can choose to view the given sum as a left, right, or midpoint Riemann sum. The choice we make determines the interval. Note that the heights of the rectangles are determined when x=1.5, 2.5, 3.5,x = 1.5,\, 2.5,\, 3.5, and 4.54.5.

Figure from prob_s1.1, line 2

Figure from prob_s1.1, line 2

  • If our sum is a right Riemann sum, then we take the heights of the rectangles from the right endpoint of each interval.

    Figure from prob_s1.1, line 2

    Figure from prob_s1.1, line 2

    Then a=0.5a=0.5 and b=4.5b=4.5. Therefore: ∑k=03f(1.5+k)⋅1\sum\limits_{k=0}^3 f (1.5 + k) \cdot 1 is a right Riemann sum on the interval [0.5,4.5][0.5,4.5] with n=4n=4.

  • If our sum is a left Riemann sum, then we take the heights of the rectangles from the left endpoint of each interval.

    Figure from prob_s1.1, line 2

    Figure from prob_s1.1, line 2

    Then a=1.5a=1.5 and b=5.5b=5.5. Therefore: ∑k=03f(1.5+k)⋅1\sum\limits_{k=0}^3 f (1.5 + k) \cdot 1 is a left Riemann sum on the interval [1.5,5.5][1.5,5.5] with n=4n=4.

  • If our sum is a midpoint Riemann sum, then we take the heights of the rectangles from the midpoint of each interval.

    Figure from prob_s1.1, line 2

    Figure from prob_s1.1, line 2

    Then a=1a=1 and b=5b=5. Therefore: ∑k=03f(1.5+k)⋅1\sum\limits_{k=0}^3 f (1.5 + k) \cdot 1 is a midpoint Riemann sum on the interval [1,5][1,5] with n=4n=4.

Q16Stage 1

Evaluate the following integral by interpreting it as a signed area, and using geometry:

∫05x dx\int_0^5 x \,\dee{x}
Hint

The area is a triangle.

Answer

252\dfrac{25}{2}

Full solution

The area in question is a triangle with base 5 and height 5, so its area is 252\dfrac{25}{2}.

Figure from prob_s1.1, line 2

Figure from prob_s1.1, line 2

Q17Stage 1

Evaluate the following integral by interpreting it as a signed area, and using geometry:

∫−25x dx\int_{-2}^5 x \,\dee{x}
Hint

There is one triangle of positive area, and one of negative area.

Answer

212\dfrac{21}{2}

Full solution

There is a positive and a negative portion of this area. The positive area is a triangle with base 5 and height 5, so area 252\dfrac{25}{2} square units. The negative area is a triangle with base 22 and height 22, so negative area 42=2\dfrac{4}{2}=2 square units. So, the net area is 252−42=212\dfrac{25}{2}-\dfrac{4}{2}=\dfrac{21}{2} square units.

Figure from prob_s1.1, line 2

Figure from prob_s1.1, line 2

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q18Stage 2Past exam · M105 2014A

Use sigma notation to write the midpoint Riemann sum for f(x)=x8f(x)=x^8 on [5,15][5,15] with n=50n=50. Do not evaluate the Riemann sum.

Hint

Review Definition 1.1.11 in the

CLP-2 text.

Answer

∑i=150(5+(i−1 ⁣/2)15)8 15\sum\limits_{i=1}^{50} \Big(5+\big(i-\nicefrac{1}{2}\big)\frac{1}{5}\Big)^8 \ \frac{1}{5}

Full solution

In general, the midpoint Riemann sum is given by

∑i=1nf(a+(i−1 ⁣/2)Δx) Δx ,where Δx=b−an.\begin{align*} \sum_{i=1}^n f\Big(a+\big(i-\nicefrac{1}{2}\big)\De x\Big)\ \De x \, , \qquad \text{where } \De x = \frac{b-a}{n}. \end{align*}

In this problem we are told that f(x)=x8f(x)=x^8, a=5a=5, b=15b=15 and n=50n=50, so that Δx=b−an=15\De x = \frac{b-a}{n} = \frac{1}{5} and the desired Riemann sum is:

∑i=150(5+(i−1 ⁣/2)15)8 15\begin{align*} \sum_{i=1}^{50} \Big(5+\big(i-\nicefrac{1}{2}\big)\frac{1}{5}\Big)^8 \ \frac{1}{5} \end{align*}
Q19Stage 2Past exam · 2016Q1

Estimate ∫−15x3  dx\displaystyle\int_{-1}^5 x^3\,\,\dee{x} using three approximating rectangles and left hand end points.

Answer

5454

Full solution

The given integral has interval of integration going from a=−1a=-1 to b=5b=5. So when we use three approximating rectangles, all of the same width, the common width is Δx=b−an=2\Delta x=\frac{b-a}{n} = 2. The first rectangle has left endpoint x0=a=−1x_0=a=-1, the second has left hand endpoint x1=a+Δx=1x_1=a+\Delta x=1, and the third has left hand end point x2=a+2Δx=3x_2=a+2\Delta x=3. So

∫−15x3  dx≈[f(x0)+f(x1)+f(x2)]Δx=[(−1)3+13+33]×2=54\begin{align*} \int_{-1}^5 x^3\,\,\dee{x} \approx \big[f(x_0)+f(x_1)+f(x_2)\big]\Delta x =\big[(-1)^3+1^3+3^3\big]\times2 =54 \end{align*}
Q20Stage 2Past exam · 2016Q1

Let ff be a function on the whole real line. Express ∫−17f(x)  dx\displaystyle\int_{-1}^{7}f(x)\,\,\dee{x} as a limit of Riemann sums, using the right endpoints.

Hint

You'll want the limit as nn goes to infinity of a sum with nn terms. If you're having a hard time coming up with the sum in terms of nn, try writing a sum with a finite number of terms of your choosing. Then, think about how that sum would change if it had nn terms.

Answer

∫−17f(x)  dx=lim⁡n→∞∑i=1nf(−1+8in)8n\displaystyle\int_{-1}^{7}f(x)\,\,\dee{x}=\displaystyle\lim_{n\to\infty}\displaystyle\sum_{i=1}^{n} f\left(-1+\frac{8i}{n}\right)\frac{8}{n}

Full solution

In the given integral, the domain of integration runs from a=−1a=-1 to b=7b=7. So, we have Δx=(b−a)n=(7−(−1))n=8n\Delta x = \frac{(b-a)}{n}= \frac{(7-(-1))}{n} = \frac{8}{n}. The left-hand end of the first subinterval is at x0=a=−1x_0=a=-1. So, the right-hand end of the ithi^{\rm th} interval is at xi∗=−1+8inx_i^* = -1+\frac{8i}{n}. So:

∫−17f(x)  dx=lim⁡n→∞∑i=1nf(−1+8in)8n\begin{align*} \int_{-1}^{7}f(x)\,\,\dee{x} =\lim_{n\to\infty}\sum_{i=1}^{n} f\left(-1+\frac{8i}{n}\right)\frac{8}{n} \end{align*}
Q21Stage 2Past exam · 2016Q1

The value of the following limit is equal to the area below a graph of y=f(x)y=f(x), integrated over the interval [0,b][0,b]:

lim⁡n→∞∑i=1n4n[sin⁡(2+4in)]2\begin{align*} \lim_{n \to \infty} \sum_{i=1}^{n} \frac{4}{n} \left[ \sin \left( 2 + \frac{4i}{n}\right)\right]^2 \end{align*}

Find f(x)f(x) and bb.

Hint

The main step is to express the given sum as the right Riemann sum,

∑i=1nf(a+iΔx)Δx.\sum_{i=1}^{n} f(a+i\De x)\Delta x.

Don't be afraid to guess Δx\De x and f(xf(x) (review Definition 1.1.11 in the

CLP-2 text). Then write out explicitly ∑i=1nf(a+iΔx)Δx\sum\limits_{i=1}^{n} f(a+i\De x)\Delta x with your guess substituted in, and compare the result with the given sum. Adjust your guess if they don't match.

Answer

f(x)=sin⁡2(2+x)f(x) = \sin^2 (2 + x) and b=4b=4

Full solution

We identify the given sum as the right Riemann sum ∑i=1nf(a+iΔx)Δx\sum\limits_{i=1}^{n} f(a+i\De x)\Delta x, with a=0a=0 (that's specified in the statement of the question). Since 4n\frac{4}{n} is multiplied in every term, and is also multiplied by ii, we let Δx=4n\Delta x = \frac{4}{n}. Then xi∗=a+iΔx=4inx_i^* = a+i\De x=\frac{4i}{n} and f(x)=sin⁡2(2+x)f(x) = \sin^2 (2 + x). So, b=a+nΔx=0+n⋅4n=4b=a+n\De x=0+n\cdot\frac{4}{n}=4.

Q22Stage 2Past exam · M105 2102A

For a certain function f(x)f(x), the following equation holds:

lim⁡n→∞∑k=1nkn21−k2n2=∫01f(x)  dx\begin{equation*} \lim_{n\rightarrow\infty}\sum\limits_{k=1}^n \frac{k}{n^2}\sqrt{1-\frac{k^2}{n^2}} =\int_0^1 f(x)\ \,\dee{x} \end{equation*}

Find f(x)f(x).

Hint

The main step is to express the given sum as the right Riemann sum ∑k=1nf(a+kΔx)Δx\sum\limits_{k=1}^{n} f(a+k\De x)\Delta x. Don't be afraid to guess Δx\De x and f(xf(x) (review Definition 1.1.11 in the

CLP-2 text). Then write out explicitly ∑k=1nf(a+kΔx)Δx\sum\limits_{k=1}^{n} f(a+k\De x)\Delta x with your guess substituted in, and compare the result with the given sum. Adjust your guess if they don't match.

Answer

f(x)=x1−x2f(x)=x\sqrt{1-x^2}

Full solution

The given sum is of the form

lim⁡n→∞∑k=1nkn21−k2n2=lim⁡n→∞∑k=1n(1n)kn1−(kn)2=lim⁡n→∞∑k=1nΔxf(xk∗)\begin{align*} \lim_{n\rightarrow\infty}\sum_{k=1}^n \frac{k}{n^2}\sqrt{1-\frac{k^2}{n^2}} =\lim_{n\rightarrow\infty}\sum_{k=1}^n \left(\frac{1}{n}\right) \frac{k}{n}\sqrt{1-\left(\frac{k}{n}\right)^2} =\lim_{n\rightarrow\infty}\sum_{k=1}^n \De x f(x_k^*) \end{align*}

with Δx=1n\De x=\frac{1}{n}, a=0a=0, xk∗=kn=a+kΔxx_k^*=\frac{k}{n}=a+k\De x and f(x)=x1−x2f(x)=x\sqrt{1-x^2}. Since x0∗=0x_0^*=0 and xn∗=1x_n^*=1, the right hand side is the definition (using the right Riemann sum) of ∫01f(x)  dx\int_0^1 f(x)\,\,\dee{x}.

Q23Stage 2Past exam · 2016Q1

Express lim⁡n→∞∑i=1n3ne−i/ncos⁡(3in)\displaystyle\lim_{n\to\infty}\displaystyle\sum_{i=1}^{n} \frac{3}{n} e^{-i/n} \cos\left(\frac{3i}{n}\right) as a definite integral.

Hint

The main step is to express the given sum in the form ∑i=1nf(xi∗)Δx\sum_{i=1}^{n} f(x_i^*)\Delta x. Don't be afraid to guess Δx\De x, xi∗x_i^* (for either a left or a right or a midpoint sum — review Definition 1.1.11 in the

CLP-2 text) and f(xf(x). Then write out explicitly ∑i=1nf(xi∗)Δx\sum_{i=1}^{n} f(x_i^*)\Delta x with your guess substituted in, and compare the result with the given sum. Adjust your guess if they don't match.

Answer

∫03e−x/3cos⁡(x)  dx\int_0^3 e^{-x/3}\cos(x)\,\,\dee{x}

Full solution

As ii ranges from 11 to nn, 3i/n3i/n range from 3/n3/n to 33 with jumps of Δx=3/n\De x=3/n, so this is

lim⁡n→∞∑i=1n3ne−i/ncos⁡(3i/n)=lim⁡n→∞∑i=1nf(xi∗)Δx=∫abf(x)  dx\begin{align*} \lim_{n\to\infty}\displaystyle\sum_{i=1}^{n} \frac{3}{n} e^{-i/n} \cos(3i/n) =\lim_{n\to\infty} \sum_{i=1}^{n} f(x_i^*)\Delta x =\int_a^b f(x)\,\,\dee{x} \end{align*}

where xi∗=3i/nx_i^* = 3i/n, f(x)=e−x/3cos⁡(x)f(x) = e^{-x/3}\cos(x), a=x0=0a=x_0=0 and b=xn=3b=x_n=3. Thus

lim⁡n→∞∑i=1n3ne−i/ncos⁡(3i/n)=∫03e−x/3cos⁡(x)  dx\begin{align*} \lim_{n\to\infty}\displaystyle\sum_{i=1}^{n} \frac{3}{n} e^{-i/n} \cos(3i/n) =\int_0^3 e^{-x/3}\cos(x)\,\,\dee{x} \end{align*}
Q24Stage 2Past exam · 2012A, 2014D

Let Rn=∑i=1niei/nn2\displaystyle R_n= \sum_{i=1}^{n} \frac{i e^{i/n}}{n^2}. Express lim⁡n→∞Rn\displaystyle\lim_{n\to\infty}R_n as a definite integral. Do not evaluate this integral.

Hint

The main step is to express the given sum in the form ∑i=1nf(xi∗)Δx\sum\limits_{i=1}^{n} f(x_i^*)\Delta x. Don't be afraid to guess Δx\De x, xi∗x_i^* (probably, based on the symbol RnR_n, assuming we have a right Riemann sum — review Definition 1.1.11 in the CLP-2 text) and f(xf(x). Then write out explicitly ∑i=1nf(xi∗)Δx\sum\limits_{i=1}^{n} f(x_i^*)\Delta x with your guess substituted in, and compare the result with the given sum. Adjust your guess if they don't match.

Answer

∫01xex  dx\displaystyle\int_0^1 x e^{x}\,\,\dee{x}

Full solution

As ii ranges from 11 to nn, the exponent in\frac{i}{n} ranges from 1n\frac{1}{n} to 11 with jumps of Δx=1n\De x=\frac{1}{n}. So let's try xi∗=inx_i^* =\frac{ i}{n}, Δx=1n\Delta x=\frac{1}{n}. Then:

Rn=∑i=1niei/nn2=∑i=1ninei/n1n=∑i=1nxi∗exi∗Δx=∑i=1nf(xi∗)Δx\begin{align*} R_n= \sum_{i=1}^{n} \frac{i e^{i/n}}{n^2} = \sum_{i=1}^{n} \frac{i}{n} e^{i/n} \frac{1}{n} = \sum_{i=1}^{n} x_i^* e^{x_i^*} \Delta x = \sum_{i=1}^{n} f(x_i^*) \Delta x \end{align*}

with f(x)=xexf(x)= x e^x, and the limit

lim⁡n→∞Rn=lim⁡n→∞∑i=1nf(xi∗)Δx=∫abf(x)  dx\begin{align*} \lim_{n\to\infty} R_n =\lim_{n\to\infty} \sum_{i=1}^{n} f(x_i^*)\Delta x =\int_a^b f(x)\,\,\dee{x} \end{align*}

Since we chose xi∗=in=0+iΔxx_i^* = \frac{i}{n} = 0+i\De x, we let a=0a=0. Then 1n=Δx=b−an=bn\frac{1}{n}=\De x = \frac{b-a}{n}=\frac{b}{n} tells us b=1b=1. Thus,

lim⁡n→∞Rn=∫01xex  dx .\begin{align*} \lim_{n\to\infty}R_n =\int_0^1 xe^{x}\,\,\dee{x}\,. \end{align*}
Q25Stage 2Past exam · 2016A

Express lim⁡n→∞(∑i=1ne−1−2i/n⋅2n)\displaystyle\lim_{n\rightarrow\infty} \bigg( \sum_{i=1}^n e^{-1-2i/n}\cdot \frac{2}{n} \bigg) as an integral in three different ways.

Hint

Try several different choices of Δx\De x and xi∗x_i^*.

Answer

Possible answers include:
∫02e−1−x  dx\displaystyle\int\limits_0^2 e^{-1-x}\ \,\dee{x}, ∫13e−x  dx\displaystyle\int\limits_1^3 e^{-x}\ \,\dee{x}, 2∫1/23/2e−2x  dx2\displaystyle\int_{1/2}^{3/2} e^{-2x}\ \,\dee{x}, and 2∫01e−1−2x  dx2\displaystyle\int\limits_0^1 e^{-1-2x}\ \,\dee{x}.

Full solution

  • If we set Δx=2n\De x = \frac{2}{n} and xi∗=2inx_i^*= \frac{2i}{n}, i.e. xi∗=a+iΔxx_i^* = a + i\De x with a=0a=0, then

    lim⁡n→∞(∑i=1ne−1−2i/n⋅2n)=lim⁡n→∞(∑i=1ne−1−xi∗Δx)=lim⁡n→∞(∑i=1nf(xi∗)Δx)with f(x)=e−1−x=∫abf(x)  dxwith a=x0=0 and b=xn=2=∫02e−1−x  dx\begin{align*} \lim_{n\rightarrow\infty} \bigg( \sum_{i=1}^n e^{-1-2i/n}\cdot \frac{2}{n} \bigg) &=\lim_{n\rightarrow\infty} \bigg( \sum_{i=1}^n e^{-1-x_i^*}\De x \bigg) \\ &=\lim_{n\rightarrow\infty} \bigg( \sum_{i=1}^n f(x_i^*)\De x \bigg) &&\text{with }f(x) = e^{-1-x} \\ &=\int_a^b f(x)\,\,\dee{x}&&\text{with }a=x_0=0\text{ and }b=x_n=2 \\ &=\int_0^2 e^{-1-x}\ \,\dee{x} \end{align*}

  • If we set Δx=2n\De x = \frac{2}{n} and xi∗=1+2inx_i^*= 1+\frac{2i}{n}, i.e. xi∗=a+iΔxx_i^* = a + i\De x with a=1a=1, then

    lim⁡n→∞(∑i=1ne−1−2i/n⋅2n)=lim⁡n→∞(∑i=1ne−xi∗Δx)=lim⁡n→∞(∑i=1nf(xi∗)Δx)with f(x)=e−x=∫abf(x)  dxwith a=x0=1 and b=xn=3=∫13e−x  dx\begin{align*} \lim_{n\rightarrow\infty} \bigg( \sum_{i=1}^n e^{-1-2i/n}\cdot \frac{2}{n} \bigg) &=\lim_{n\rightarrow\infty} \bigg( \sum_{i=1}^n e^{-x_i^*}\De x \bigg) \\ &=\lim_{n\rightarrow\infty} \bigg( \sum_{i=1}^n f(x_i^*)\De x \bigg) &&\text{with }f(x) = e^{-x} \\ &=\int_a^b f(x)\,\,\dee{x}&&\text{with }a=x_0=1\text{ and }b=x_n=3 \\ &=\int_1^3 e^{-x}\ \,\dee{x} \end{align*}

  • If we set Δx=1n\De x = \frac{1}{n} and xi∗=inx_i^*= \frac{i}{n}, i.e. xi∗=a+iΔxx_i^* = a + i\De x with a=0a=0, then

    lim⁡n→∞(∑i=1ne−1−2i/n⋅2n)=lim⁡n→∞(∑i=1ne−1−2xi∗ 2Δx)=lim⁡n→∞(∑i=1nf(xi∗)Δx)with f(x)=2e−1−2x=∫abf(x)  dxwith a=x0=0 and b=xn=1=2∫01e−1−2x  dx\begin{align*} \lim_{n\rightarrow\infty} \bigg( \sum_{i=1}^n e^{-1-2i/n}\cdot \frac{2}{n} \bigg) &=\lim_{n\rightarrow\infty} \bigg( \sum_{i=1}^n e^{-1-2x_i^*}\ 2\De x \bigg) \\ &=\lim_{n\rightarrow\infty} \bigg( \sum_{i=1}^n f(x_i^*)\De x \bigg) &&\text{with }f(x) = 2e^{-1-2x} \\ &=\int_a^b f(x)\,\,\dee{x}&&\text{with }a=x_0=0\text{ and }b=x_n=1 \\ &=2\int_0^1 e^{-1-2x}\ \,\dee{x} \end{align*}

  • If we set Δx=1n\De x = \frac{1}{n} and xi∗=12+inx_i^*= \frac{1}{2}+\frac{i}{n}, i.e. xi=a+iΔxx_i = a + i\De x with a=12a=\frac{1}{2}, then

    lim⁡n→∞(∑i=1ne−1−2i/n⋅2n)=lim⁡n→∞(∑i=1ne−2xi 2Δx)=lim⁡n→∞(∑i=1nf(xi∗)Δx)with f(x)=2e−2x=∫abf(x)  dxwith a=x0=12 and b=xn=32=2∫1/23/2e−2x  dx\begin{align*} \lim_{n\rightarrow\infty} \bigg( \sum_{i=1}^n e^{-1-2i/n}\cdot \frac{2}{n} \bigg) &=\lim_{n\rightarrow\infty} \bigg( \sum_{i=1}^n e^{-2x_i}\ 2\De x \bigg) \\ &=\lim_{n\rightarrow\infty} \bigg( \sum_{i=1}^n f(x_i^*)\De x \bigg) &&\text{with }f(x) = 2e^{-2x} \\ &=\int_a^b f(x)\,\,\dee{x}&&\text{with }a=x_0=\frac{1}{2}\text{ and }b=x_n=\frac{3}{2} \\ &=2\int_{1/2}^{3/2} e^{-2x}\ \,\dee{x} \end{align*}

Questions 26 and 27 use the formula for a geometric sum, Equation 1.1.3 in the CLP-2 text.

Q26Stage 2

Evaluate the sum 1+r3+r6+r9+⋯+r3n1+r^3+r^6+r^9+\cdots+r^{3n}.

Hint

Let x=r3x=r^3, and re–write the sum in terms of xx.

Answer

r3n+3−1r3−1\dfrac{r^{3n+3}-1}{r^3-1}

Full solution

This is similar to the familiar form of a geometric sum, but the powers go up by threes. So, we make a subsitution. If x=r3x=r^3, then:

1+r3+r6+r9+⋯+r3n=1+x+x2+x3+⋯+xn1+r^3+r^6+r^9+\cdots+r^{3n}=1+x+x^2+x^3+\cdots+x^n

Now, using Equation 1.1.3 in the CLP-2 text,

1+x+x2+x3+⋯+xn=xn+1−1x−11+x+x^2+x^3+\cdots+x^n = \frac{x^{n+1}-1}{x-1}

Substituting back in x=r3x=r^3, we find our sum is equal to (r3)n+1−1r3−1\dfrac{(r^3)^{n+1}-1}{r^3-1}, or r3n+3−1r3−1\dfrac{r^{3n+3}-1}{r^3-1}.

Q27Stage 2

Evaluate the sum r5+r6+r7+⋯+r100r^5+r^6+r^7+\cdots+r^{100}.

Hint

Note the sum does not start at r0=1r^0=1.

Answer

r5(r96−1r−1)r^5\left(\dfrac{r^{96}-1}{r-1}\right)

Full solution

The sum does not start at 11, so we need to do some algebra. We can either factor out the first term, or subtract off the initial terms that are missing.

  • If we factor out r5r^5, then what's left fits the form of Equation 1.1.3 in the CLP-2 text:

    r5+r6+r7+⋯+r100=r5[1+r+r2+⋯+r95]=r5(r96−1r−1) .r^5+r^6+r^7+\cdots+r^{100}=r^5\left[1+r+r^2+\cdots + r^{95}\right]= r^5\left(\frac{r^{96}-1}{r-1}\right)\ .

  • We know how to evaluate sums of this form if they start at 1, so we re-write our sum as follows:

    r5+r6+r7+⋯+r100=(1+r+r2+r3+r4+r5+⋯+r100)−(1+r+r2+r3+r4)=r101−1r−1−r5−1r−1=r101−1−r5+1r−1=r101−r5r−1=r5(r96−1r−1) .\begin{align*} r^5+r^6+r^7+\cdots+r^{100}&=\left(1+r+r^2+r^3+r^4+r^5+\cdots+r^{100}\right) - \left(1+r+r^2+r^3+r^4\right) \\ &=\frac{r^{101}-1}{r-1} - \frac{r^5-1}{r-1}\\ &=\frac{r^{101}-1-r^5+1}{r-1}=\frac{r^{101}-r^5}{r-1}=r^5\left(\frac{r^{96}-1}{r-1}\right)\ . \end{align*}

Remember that a definite integral is a signed area between a curve and the xx-axis. We'll spend a lot of time learning strategies for evaluating definite integrals, but we already know lots of ways to find area of geometric shapes. In Questions 28 through 33, use your knowledge of geometry to find the signed areas described by the integrals given.

Q28Stage 2Past exam · M105 2013A

Evaluate ∫−12∣2x∣  dx{\displaystyle\int_{-1}^2 |2x|\ \,\dee{x}}.

Hint

Draw a picture. See Example 1.1.15 in the

CLP-2 text.

Answer

55

Full solution

Recall that

∣x∣={−xif x≤0xif x≥0\begin{align*} |x|=\begin{cases} -x &\text{if }x\le 0\\ x &\text{if }x\ge 0 \end{cases} \end{align*}

so that

∣2x∣={−2xif x≤02xif x≥0\begin{align*} |2x|=\begin{cases} -2x &\text{if }x\le 0\\ 2x &\text{if }x\ge 0 \end{cases} \end{align*}

To picture the geometric figure whose area the integral represents observe that

  • at the left hand end of the domain of integration x=−1x=-1 and the integrand ∣2x∣=∣−2∣=2|2x|=|-2|=2 and

  • as xx increases from −1-1 towards 00, the integrand ∣2x∣=−2x|2x|=-2x decreases linearly, until

  • when xx hits 00 the integrand hits ∣2x∣=∣0∣=0|2x|=|0|=0 and then

  • as xx increases from 00, the integrand ∣2x∣=2x|2x|=2x increases linearly, until

  • when xx hits +2+2, the right hand end of the domain of integration, the integrand hits ∣2x∣=∣4∣=4|2x|=|4|=4.

So the integral ∫−12∣2x∣  dx\int_{-1}^2 |2x|\ \,\dee{x} is the area of the union of the two shaded triangles (one of base 11 and of height 22 and the other of base 22 and height 44) in the figure on the right below and

∫−12∣2x∣  dx=12×1×2+12×2×4=5\includegraphics\begin{align*} \int_{-1}^2 |2x|\ \,\dee{x} = \frac{1}{2}\times 1\times 2 + \frac{1}{2}\times 2\times 4 = 5\qquad\qquad {\includegraphics{OEM105_13A_1e}} \end{align*}
Q29Stage 2

Evaluate the following integral by interpreting it as a signed area, and using geometry:

∫−35∣t−1∣ dt\int_{-3}^5 |t-1| \,\dee{t}
Hint

Draw a picture. Remember ∣x∣={xx≥0−xx<0|x| = \left\{\begin{array}{rc}x&x\ge 0\\-x&x<0\end{array}\right. .

Answer

16

Full solution

The area we want is two triangles, both above the xx-axis. Each triangle has base 44 and height 44, so the total area is 2⋅(4⋅42)=162\cdot\left(\dfrac{4\cdot 4}{2}\right)=16.

Figure from prob_s1.1, line 2

Figure from prob_s1.1, line 2

If you had a hard time sketching the function, recall that the absolute value of a number leaves it unchanged if it is positive or zero, and flips the sign if it is negative. So, when t−1≥0t-1 \ge 0 (that is, when t≥1t \ge 1), our function is simply f(t)=∣t−1∣=t−1f(t)=|t-1|=t-1. On the other hand, when t=1t=1 is negative (that is, when t<1t<1), the absolute value changes the sign, so f(t)=∣t−1∣=−(t−1)=−t+1f(t) = |t-1|=-(t-1)=-t+1.

Q30Stage 2

Evaluate the following integral by interpreting it as a signed area, and using geometry:

∫abx dx\int_a^b x \,\dee{x}

where 0≤a≤b0 \leq a \leq b.

Hint

Draw a picture: the area we want is a trapezoid. If you don't remember a formula for the area of a trapezoid, think of it as the difference of two triangles.

Answer

b2−a22\dfrac{b^2-a^2}{2}

Full solution

The area we want is a trapezoid with base (b−a)(b-a) and heights aa and bb, so its area is (b−a)(b+a)2=b2−a22\dfrac{(b-a)(b+a)}{2}=\dfrac{b^2-a^2}{2}.

Figure from prob_s1.1, line 2

Figure from prob_s1.1, line 2

Instead of using a formula for the area of a trapezoid, you can find the blue area as the area of a triangle with base and height bb, minus the area of a triangle with base and height aa.

Q31Stage 2

Evaluate the following integral by interpreting it as a signed area, and using geometry:

∫abx dx\int_a^b x\, \dee{x}

where a≤b≤0a \leq b \leq 0.

Hint

You can draw a very similar picture to Question 30, but remember the areas are negative.

Answer

b2−a22\dfrac{b^2-a^2}{2}

Full solution

The area is negative. The shape is a trapezoid with base length (b−a)(b - a) and heights 0−a=−a0-a=-a and 0−b=−b0-b=-b (note: those are nonnegative numbers), so its area is (b−a)(−b−a)2=−b2+a22\dfrac{(b-a)(-b-a)}{2}=\dfrac{-b^2+a^2}{2}. Since the shape is below the xx-axis, we change its sign. Thus, the integral evaluates to b2−a22\dfrac{b^2-a^2}{2}.

Figure from prob_s1.1, line 2

Figure from prob_s1.1, line 2

The signs can be a little hard to keep track of. The base of our trapezoid is ∣a−b∣|a-b|; since b>ab>a, this is b−ab-a. The heights of the trapezoid are ∣a∣|a| and ∣b∣|b|; since these are both negative, ∣a∣=−a|a|=-a and ∣b∣=−b|b|=-b.

We note that this is the same result as in Question 30.

Q32Stage 2

Evaluate the following integral by interpreting it as a signed area, and using geometry:

∫0416−x2 dx\int_0^4 \sqrt{16-x^2} \ \dee{x}
Hint

If y=16−x2y=\sqrt{16-x^2}, then yy is nonnegative, and y2+x2=16y^2+x^2=16.

Answer

4π4\pi

Full solution

If y=16−x2y=\sqrt{16-x^2}, then yy is nonnegative, and y2+x2=16y^2+x^2=16. So, the graph y=16−x2y=\sqrt{16-x^2} is the upper half of a circle of radius 4. Since xx only runs from 0 to 4, we have a quarter of a circle of radius 4. Then the area under the curve is 14[π⋅42]=4π\dfrac{1}{4}\left[\pi\cdot 4^2\right]=4\pi.

Figure from prob_s1.1, line 2

Figure from prob_s1.1, line 2

Q33Stage 2Past exam · 2016Q1

Use elementary geometry to calculate ∫03f(x)  dx\displaystyle \int_0^3 f(x)\,\,\dee{x}, where

f(x)={x,if x≤1,1,if x>1.\begin{align*} f(x) = \begin{cases} x, & \text{if } x \le 1,\\ 1, & \text{if } x > 1. \end{cases} \end{align*}
Hint

Sketch the graph of f(x)f(x).

Answer

∫03f(x)  dx=2.5\displaystyle\int_0^3 f(x)\,\,\dee{x} = 2.5

Full solution

Here is a sketch the graph of f(x)f(x).

Figure from prob_s1.1, line 1799

Figure from prob_s1.1, line 1799

There is a linear increase from x=0x=0 to x=1x=1, followed by a constant. Using the interpretation of ∫03f(x)  dx\int_0^3 f(x)\,\,\dee{x} as the area between y=f(x)y=f(x) and the xx–axis with xx between 00 and 33, we can break this area into:

  • ∫01f(x)  dx\int_0^1 f(x)\,\,\dee{x}: a right-angled triangle of height 11 and base 11 and hence area 0.50.5.

  • ∫13f(x)  dx\int_1^3 f(x)\,\,\dee{x}: a rectangle of height 11 and base 22 and hence area 22.

Summing up: ∫03f(x)  dx=2.5\int_0^3 f(x)\,\,\dee{x} = 2.5.

Q34Stage 2Past exam · 2016Q1

A car's gas pedal is applied at t=0t=0 seconds and the car accelerates continuously until t=2t=2 seconds. The car's speed at half-second intervals is given in the table below. Find the best possible upper estimate for the distance that the car traveled during these two seconds.

!\vrule width 1ptc!\vrule width 1ptc|c|c|c|c!\vrule width 1pt \noalign\hrule height 1pt tt (s)000.50.51.01.01.51.522
vv (m/s)014223040
\noalign\hrule height 1pt
Hint

At which time in the interval, for example, 0≤t≤0.50\le t\le 0.5, is the car moving the fastest?

Answer

53 m

Full solution

The car's speed increases with time. So its highest speed on any time interval occurs at the right hand end of the interval and the best possible upper estimate for the distance traveled is given by the right Riemann sum with Δx=0.5\Delta x =0.5, which is

[v(0.5)+v(1.0)+v(1.5)+v(2.0)]×0.5=[14+22+30+40]×0.5=53 m\begin{equation*} \big[v(0.5)+v(1.0)+v(1.5)+v(2.0)\big]\times 0.5 =\big[14+22+30+40\big]\times 0.5= 53\text{ m} \end{equation*}
Q35Stage 2

True or false: the answer you gave for Question 34 is definitely greater than or equal to the distance the car travelled during the two seconds in question.

Hint

What are the possible speeds the car could have reached at time t=0.25t=0.25?

Answer

true

Full solution

There is a key detail in the statement of Question 34: namely, that the car is continuously accelerating. So, although we don't know exactly what's going on in between our brief snippets of information, we know that the car is not going any faster during an interval than at the end of that interval. Therefore, the car certainly travelled no farther than our estimation.

We ask this question in order to point out an important detail. If we did not have the information that the car was continuously accelerating, we would not be able to give a certain upper bound on its distance travelled. It would be possible that, when the car is not being observed (for example, when t=0.25t=0.25), it is going much faster than when it is being observed.

Q36Stage 2

An airplane's speed at one-hour intervals is given in the table below. Approximate the distance travelled by the airplane from noon to 4pm using a midpoint Riemann sum.

!\vrule width 1ptc!\vrule width 1ptc|c|c|c|c!\vrule width 1pt \noalign\hrule height 1pt time12:00 pm1:00 pm2:00 pm3:00 pm4:00 pm
speed (km/hr)800700850900750
\noalign\hrule height 1pt
Hint

You need to know the speed of the plane at the midpoints of your intervals, so (for example) noon to 1pm is not one of your intervals.

Answer

3200 km

Full solution

First, note that the distance travelled by the plane is equal to the area under the graph of its speed.

We need to know the speed of the plane at the midpoints of our intervals. So (for example) noon to 1pm is not one of your intervals–we don't know the speed at 12:30. (A common idea is to average the two end values, 700 and 800. This is a fine approximation, but it is not a Riemann sum.) So, we use the two intervals 12:00 to 2:00, and 2:00 to 4:00. Then our intervals have length 2 hours, and at the midpoints of the intervals the speed of the plane is 700 kph and 900 kph, respectively. So, our midpoint Riemann sum gives us:

700(2)+900(2)=3200700(2)+900(2) = 3200

an approximation of 3200 km travelled by the plane from noon to 4:00 pm.

Remark: if we had been asked to approximate the distance travelled from 11:30 am to 4:30 pm, then we could have used the midpoint rule with five intervals and made use of every entry in the data table. With the question as stated, however, we ignore three out of five entries in the table because they are not the midpoints of our intervals.

3Stage 3Application

Further than practice: several ideas at once, or an unfamiliar situation.

Q37Stage 3Past exam · 2016Q1

(a) Express

lim⁡n→∞∑i=1n2n4−(−2+2in)2\begin{equation*} \lim_{n\rightarrow\infty} \sum_{i=1}^n\frac{2}{n}\sqrt{4-\left(-2+\frac{2i}{n}\right)^2} \end{equation*}

as a definite integal.

(b) Evaluate the integral of part (a).

Hint

Sure looks like a Riemann sum.

Answer

(a) There are many possible answers. Two are ∫−204−x2  dx\int_{-2}^0 \sqrt{4-x^2}\,\,\dee{x} and ∫024−(−2+x)2  dx\int_0^2 \sqrt{4-(-2+x)^2}\,\,\dee{x}. (b) π\pi

Full solution

  • Set xi∗=−2+2inx_i^*=-2+\frac{2i}{n}. Then a=x0=−2a=x_0=-2 and b=xn=0b=x_n=0 and Δx=2n\Delta x=\frac{2}{n}. So

    lim⁡n→∞∑i=1n2n4−(−2+2in)2=lim⁡n→∞∑i=1nf(xi∗)Δx with f(x)=4−x2 and Δx=2n=∫−204−x2  dx\begin{align*} \lim_{n\rightarrow\infty} \sum_{i=1}^n\frac{2}{n}\sqrt{4-\left(-2+\frac{2i}{n}\right)^2} &= \lim_{n\rightarrow\infty} \sum_{i=1}^n f(x_i^*)\Delta x\qquad \text{ with }f(x) = \sqrt{4-x^2}\text{ and }\Delta x = \frac{2}{n} \\ &=\int_{-2}^0 \sqrt{4-x^2}\,\,\dee{x} \end{align*}

    For the integral ∫−204−x2  dx\int_{-2}^0 \sqrt{4-x^2}\,\,\dee{x}, y=4−x2y=\sqrt{4-x^2} is equivalent to x2+y2=4x^2+y^2=4, y≥0y\ge 0. So the integral represents the area between the upper half of the circle x2+y2=4x^2+y^2=4 (which has radius 22) and the xx-axis with −2≤x≤0-2\le x\le 0, which is a quarter circle with area 14⋅π 22=π\frac{1}{4}\cdot \pi\, 2^2 = \pi.

    Figure from prob_s1.1, line 2

    Figure from prob_s1.1, line 2

  • Set xi∗=2inx_i^*=\frac{2i}{n}. Then a=x0=0a=x_0=0 and b=xn=2b=x_n=2 and Δx=2n\Delta x=\frac{2}{n}. So

    lim⁡n→∞∑i=1n2n4−(−2+2in)2=lim⁡n→∞∑i=1nf(xi∗)Δx with f(x)=4−(−2+x)2, Δx=2n=∫024−(−2+x)2  dx\begin{align*} \lim_{n\rightarrow\infty} \sum_{i=1}^n\frac{2}{n}\sqrt{4-\left(-2+\frac{2i}{n}\right)^2} &= \lim_{n\rightarrow\infty} \sum_{i=1}^n f(x_i^*)\Delta x\quad \text{ with }f(x) = \sqrt{4-(-2+x)^2}\text{, }\Delta x = \frac{2}{n} \\ &=\int_0^2 \sqrt{4-(-2+x)^2}\,\,\dee{x} \end{align*}

    For the integral ∫024−(−2+x)2  dx\int_{0}^2 \sqrt{4-(-2+x)^2}\,\,\dee{x} , y=4−(x−2)2y=\sqrt{4-(x-2)^2} is equivalent to (x−2)2+y2=4(x-2)^2+y^2=4, y≥0y\ge 0. So the integral represents the area between the upper half of the circle (x−2)2+y2=4(x-2)^2+y^2=4 (which is centered at (2,0)(2,0) and has radius 22) and the xx-axis with 0≤x≤20\le x\le 2, which is a quarter circle with area 14⋅π 22=π\frac{1}{4}\cdot \pi\, 2^2 = \pi.

    Figure from prob_s1.1, line 2

    Figure from prob_s1.1, line 2

Q38Stage 3Past exam · 2016Q1

Consider the integral:

∫03(7+x3)  dx.         (∗)\begin{align*} \int_0^3 (7 + x^3) \,\,\dee{x} . ~~~~~~~~~ (*) \end{align*}

  1. Approximate this integral using the left Riemann sum with n=3n=3 intervals.

  2. Write down the expression for the right Riemann sum with nn intervals and calculate the sum. Now take the limit n→∞n \to \infty in your expression for the Riemann sum, to evaluate the integral (∗*) exactly.

You may use the identity

∑i=1ni3=n4+2n3+n24\begin{align*} \sum_{i=1}^{n} i^3 = \frac{n^4 +2n^3 + n^2}{4} \end{align*}
Hint

For part (b): don't panic! Just take it one step at a time. The first step is to write down the Riemann sum. The second step is to evaluate the sum, using the given identity. The third step is to evaluate the limit n→∞n\rightarrow\infty.

Answer

(a) 3030 (b) 411441 \frac{1}{4}

Full solution

(a) The left Riemann sum is defined as

Ln=∑i=1nf(xi−1)Δxwith xi=a+iΔx\begin{align*} L_n = \sum_{i=1}^{n} f(x_{i-1})\Delta x \qquad\text{with }x_i=a+i\De x \end{align*}

We subdivide into n=3n=3 intervals, so that Δx=b−an=3−03=1\Delta x = \frac{b-a}{n} =\frac{3-0}{3}=1, x0=0x_0=0, x1=1x_1=1 and x2=2x_2=2. The function f(x)=7+x3f(x) = 7 + x^3 has the values f(x0)=7+03=7f(x_0) = 7+0^3=7, f(x1)=7+13=8f(x_1) = 7+1^3=8, and f(x2)=7+23=15f(x_2) = 7+2^3=15, from which we evaluate

L3=[f(x0)+f(x1)+f(x2)]Δx=[7+8+15]×1=30\begin{align*} L_3 = \big[f(x_0)+f(x_1)+f(x_2)\big]\De x = \big[7+8+15\big]\times 1= 30 \end{align*}

(b) We divide into nn intervals so that Δx=b−an=3n\Delta x = \frac{b-a}{n}=\frac{3}{n} and xi=a+iΔx=3inx_i = a+i\De x= \frac{3i}{n}. The right Riemann sum is therefore:

Rn=∑i=1nf(xi)Δx=∑i=1n[7+(3i)3n3]3n=∑i=1n[21n+81 i3n4]\begin{align*} R_n = \sum_{i=1}^{n} f(x_i)\Delta x = \sum_{i=1}^{n} \left[ 7 + \frac{(3i)^3}{n^3} \right] \frac{3}{n} = \sum_{i=1}^{n} \left[ \frac{21}{n} + \frac{81\,i^3}{n^4} \right] \end{align*}

To calculate the sum:

Rn=(21n∑i=1n1)+(81n4∑i=1ni3)=(21n×n)+(81n4×n4+2n3+n24)=21+814(1+2/n+1/n2)\begin{align*} R_n &= \left( \frac{21}{n} \sum_{i=1}^{n} 1 \right) + \left( \frac{81}{n^4} \sum_{i=1}^{n} i^3 \right)\\ &=\left( \frac{21}{n}\times n \right)+\left( \frac{81}{n^4} \times \frac{n^4 +2n^3 + n^2}{4} \right)\\ &= 21 + \frac{81}{4}(1 +2/n + 1/n^2) \end{align*}

To evaluate the limit exactly, we take n→∞n \to \infty. The expressions involving 1/n1/n vanish leaving:

∫03(7+x3)  dx=lim⁡n→∞Rn=21+814=4114\begin{align*} \int_0^3 (7 + x^3) \,\,\dee{x} = \lim_{n \to \infty} R_n = 21 + \frac{81}{4} = 41 \frac{1}{4} \end{align*}
Q39Stage 3Past exam · 2013A

Using a limit of right–endpoint Riemann sums, evaluate ∫24x2  dx\displaystyle\int_2^4 x^2\ \,\dee{x}.

You may use the formulas ∑i=1ni=n(n+1)2\sum\limits_{i=1}^n i = \frac{n(n + 1)}{2} and ∑i=1ni2=n(n+1)(2n+1)6\sum\limits_{i=1}^n i^2 = \frac{n(n + 1)(2n + 1)}{6}.

Hint

The first step is to write down the Riemann sum. The second step is to evaluate the sum, using the given formulas. The third step is to evaluate the limit as n→∞n\rightarrow\infty.

Answer

563\dfrac{56}{3}

Full solution

In general, the right–endpoint Riemann sum approximation to the integral ∫abf(x)  dx\int_a^b f(x)\,\,\dee{x} using nn rectangles is

∑i=1nf(a+iΔx)Δx\begin{align*} \sum_{i=1}^n f(a+i\De x) \De x \end{align*}

where Δx=b−an\De x=\frac{b-a}{n}. In this problem, a=2a=2, b=4b=4, and f(x)=x2f(x)=x^2, so that Δx=2n\De x=\frac{2}{n} and the right–endpoint Riemann sum approximation becomes

∑i=1nf(2+2in)2n=∑i=1n(2+2in)22n=∑i=1n(4+8in+4i2n2)2n=∑i=1n(8n+16in2+8i2n3)=∑i=1n8n+∑i=1n16in2+∑i=1n8i2n3=8n∑i=1n1+16n2∑i=1ni+8n3∑i=1ni2=8nn+16n2⋅n(n+1)2+8n3⋅n(n+1)(2n+1)6=8+8(1+1n)+43(1+1n)(2+1n)\begin{align*} \sum_{i=1}^n f\Big(2+\frac{2i}{n}\Big) \frac{2}{n}&= \sum_{i=1}^n \Big(2+\frac{2i}{n}\Big)^2 \frac{2}{n}\\ &=\sum_{i=1}^n \left(4+\frac{8i}{n}+\frac{4i^2}{n^2}\right)\frac{2}{n} \\&=\sum_{i=1}^n \left(\frac{8}{n}+\frac{16i}{n^2}+\frac{8i^2}{n^3}\right) \\&=\sum_{i=1}^n \frac{8}{n}+\sum_{i=1}^n \frac{16i}{n^2} +\sum_{i=1}^n \frac{8i^2}{n^3} \cr &=\frac{8}{n}\sum_{i=1}^n 1+\frac{16}{n^2}\sum_{i=1}^n i +\frac{8}{n^3}\sum_{i=1}^n i^2\cr &=\frac{8}{n}n + \frac{16}{n^2}\cdot\frac{n(n+1)}{2} +\frac{8}{n^3}\cdot\frac{n(n + 1)(2n + 1)}{6} \cr &=8 + 8\Big(1+\frac{1}{n}\Big) +\frac{4}{3}\Big(1+\frac{1}{n}\Big)\Big(2+\frac{1}{n}\Big) \cr \end{align*}

So

∫24x2  dx=lim⁡n→∞[8+8(1+1n)+43(1+1n)(2+1n)]=8+8+43×2=563\begin{align*} \int_2^4 x^2\ \,\dee{x}=\lim_{n\rightarrow\infty} \Big[8 + 8\Big(1+\frac{1}{n}\Big) +\frac{4}{3}\Big(1+\frac{1}{n}\Big) \big(2+\frac{1}{n}\Big)\Big] =8+8+\frac{4}{3}\times 2 =\frac{56}{3} \end{align*}
Q40Stage 3Past exam · 2016Q1

Find ∫02(x3+x)  dx\displaystyle\int_0^2 (x^3+x)\,\,\dee{x} using the definition of the definite integral. You may use the summation formulas ∑i=1ni3=n4+2n3+n24\sum\limits_{i=1}^{n}i^3 = \frac{n^4+2n^3+n^2}4 and ∑i=1ni=n2+n2\sum\limits_{i=1}^{n} i = \frac{n^2+n}{2}.

Hint

The first step is to write down the Riemann sum. The second step is to evaluate the sum, using the given formulas. The third step is to evaluate the limit n→∞n\rightarrow\infty.

Answer

66

Full solution

We'll use right Riemann sums with a=0a=0 and b=2b=2. When there are nn rectangles, Δx=b−an=2n\Delta x = \frac{b-a}{n}=\frac{2}{n} and xi=a+iΔx=2i/nx_i = a+i\De x=2i/n. So we need to evaluate

lim⁡n→∞∑i=1nf(xi)Δx=lim⁡n→∞∑i=1n((xi)3+xi)Δx=lim⁡n→∞∑i=1n((2in)3+2in)2n=lim⁡n→∞2n∑i=1n(8i3n3+2in)=lim⁡n→∞(16n4∑i=1ni3+4n2∑i=1ni)=lim⁡n→∞(16(n4+2n3+n2)n4⋅4+4(n2+n)n2⋅2)=lim⁡n→∞(164(1+2n+1n2)+42(1+1n))=164+42=6.\begin{align*} \lim_{n\to\infty} \sum_{i=1}^{n} f(x_i)\De x &=\lim_{n\to\infty} \sum_{i=1}^{n} \left( (x_i)^3 + x_i\right) \De x\\ &=\lim_{n\to\infty} \sum_{i=1}^{n} \left( \left(\frac{2i}{n}\right)^3 + \frac{2i}{n}\right) \frac{2}{n} \\ & = \lim_{n\to\infty} \frac{2}{n} \sum_{i=1}^{n} \left(\frac{8i^3}{n^3} + \frac{2i}{n}\right) \\ &= \lim_{n\to\infty} \left( \frac{16}{n^4} \sum_{i=1}^{n} i^3 + \frac{4}{n^2} \sum_{i=1}^{n} i \right) \\ & = \lim_{n\to\infty} \left( \frac{16(n^4+2n^3+n^2)}{n^4 \cdot 4} + \frac{4(n^2+n)}{n^2 \cdot 2} \right) \\ &= \lim_{n\to\infty} \left( \frac{16}{4}\left(1+\frac{2}{n} + \frac1{n^2} \right) + \frac{4}{2}\left(1+\frac{1}{n}\right)\right) \\ & = \frac{16}{4} + \frac{4}{2} = 6. \end{align*}
Q41Stage 3Past exam · 2014D

Using a limit of right–endpoint Riemann sums, evaluate ∫14(2x−1)  dx\displaystyle\int_1^4 (2x-1)\,\,\dee{x}. Do not use anti-differentiation, except to check your answer. (You'll learn about this method starting in Section 1.3 of the CLP-2 text. You can also check this answer using geometry.) You may use the formula ∑i=1ni=n(n+1)2\sum\limits_{i=1}^{n} i = \frac{n(n+1)}{2}.

Hint

You've probably seen this hint before. It is worth repeating. Don't panic! Just take it one step at a time. The first step is to write down the Riemann sum. The second step is to evaluate the sum, using the given formula. The third step is to evaluate the limit n→∞n\rightarrow\infty.

Answer

1212

Full solution

We'll use right Riemann sums with a=1a=1, b=4b=4 and f(x)=2x−1f(x) =2x-1. When there are nn rectangles, Δx=b−an=3n\Delta x = \frac{b-a}{n}=\frac{3}{n} and xi=a+iΔx=1+3i/nx_i = a+i\De x=1 + 3i/n. So we need to evaluate

lim⁡n→∞∑i=1nf(xi)Δx=lim⁡n→∞∑i=1n(2xi−1)Δx=lim⁡n→∞∑i=1n(2+6in−1)3n=lim⁡n→∞3n∑i=1n(6in+1)=lim⁡n→∞(18n2∑i=1ni+3n∑i=1n1)=lim⁡n→∞(18⋅n(n+1)n2⋅2+3nn)=lim⁡n→∞(9(1+1n)+3)=9+3=12.\begin{align*} \lim_{n\to\infty} \sum_{i=1}^{n} f(x_i) \De x & = \lim_{n\to\infty} \sum_{i=1}^{n} \left( 2x_i - 1\right) \De x\\ & =\lim_{n\to\infty} \sum_{i=1}^{n} \left(2+ \frac{6i}{n}-1\right)\frac{3}{n} \\ & = \lim_{n\to\infty} \frac{3}{n} \sum_{i=1}^{n} \left(\frac{6i}{n}+1\right) \\ &= \lim_{n\to\infty} \left( \frac{18}{n^2} \sum_{i=1}^{n} i + \frac{3}{n} \sum_{i=1}^{n} 1 \right) \\ & = \lim_{n\to\infty} \left(\frac{18\cdot n(n+1)}{n^2\cdot 2} + \frac{3}{n}n\right)\\ &= \lim_{n\to\infty} \left(9\left(1+\frac{1}{n}\right) + 3\right) \\ & = 9 + 3 = 12. \end{align*}
Q42Stage 3

Give a function f(x)f(x) that has the following expression as a right Riemann sum when n=10n=10, Δ(x)=10\Delta(x)=10 and a=−5a=-5:

∑i=1103(7+2i)2sin⁡(4i) .\sum_{i=1}^{10} 3(7+2i)^2\sin(4i)\,.
Hint

Using the definition of a right Riemann sum, we can come up with an expression for f(−5+10i)f(-5+10i). In order to find f(x)f(x), set x=−5+10ix=-5+10i.

Answer

f(x)=310(x5+8)2sin⁡(2x5+2)f(x)=\dfrac{3}{10}\left(\dfrac{x}{5}+8\right)^2\sin\left(\dfrac{2x}{5}+2\right)

Full solution

Using the definition of a right Riemann sum,

∑i=1103(7+2i)2sin⁡(4i)=∑i=110Δxf(a+iΔx)\begin{align*}\displaystyle\sum_{i=1}^{10} 3(7+2i)^2\sin(4i) &= \displaystyle\sum_{i=1}^{10} \Delta x f(a+i\Delta x)\end{align*}

Since Δx=10\Delta x = 10 and a=−5a=-5,

∑i=1103(7+2i)2sin⁡(4i)=∑i=11010f(−5+10i)\begin{align*}\displaystyle\sum_{i=1}^{10} 3(7+2i)^2\sin(4i) &= \displaystyle\sum_{i=1}^{10} 10 f(-5+10i)\end{align*}

Dividing both expressions by 10,

∑i=110310(7+2i)2sin⁡(4i)=∑i=110f(−5+10i)\begin{align*}\displaystyle\sum_{i=1}^{10} \frac{3}{10}(7+2i)^2\sin(4i) &= \displaystyle\sum_{i=1}^{10} f(-5+10i)\end{align*}

So, we have an expression for f(−5+10i)f(-5+10i):

f(−5+10i)=310(7+2i)2sin⁡(4i)\begin{align*}f(-5+10i) &= \frac{3}{10}(7+2i)^2\sin(4i)\end{align*}

In order to find f(x)f(x), let x=−5+10ix=-5+10i. Then i=x10+12i=\frac{x}{10}+\frac{1}{2}.

f(x)=310(7+2(x10+12))2sin⁡(4(x10+12))=310(x5+8)2sin⁡(2x5+2) .\begin{align*}f(x) &= \frac{3}{10}\left(7+2\left(\frac{x}{10}+\frac{1}{2}\right)\right)^2\sin\left(4\left(\frac{x}{10}+\frac{1}{2}\right)\right)\\ &=\frac{3}{10}\left(\frac{x}{5}+8\right)^2\sin\left(\frac{2x}{5}+2\right)\ .\end{align*}
Q43Stage 3

Using the method of Example 1.1.2 in the CLP-2 text, evaluate

∫012x dx\int_0^1 2^x \ \dee{x}
Hint

Recall that for a positive constant aa, ddx{ax}=axlog⁡a\diff{}{x}\left\{a^x\right\} = a^x \log a, where log⁡a\log a is the natural logarithm (base ee) of aa.

Answer

1log⁡2\dfrac{1}{\log 2}

Full solution

As in the text, we'll set up a Riemann sum for the given integral. Right Riemann sums have the simplest form, so we use a right Riemann sum, but we could equally well use left or midpoint.

∫012xdx=lim⁡n→∞∑i=1nΔxf(a+iΔx)=lim⁡n→∞∑i=1n1nf(in)=lim⁡n→∞∑i=1n1n⋅2i/n=lim⁡n→∞1n(21/n+22/n+23/n+⋯+2n/n)=lim⁡n→∞21/nn(1+21/n+22/n+⋯+2n−1n)=lim⁡n→∞21/nn(1+21/n+(21/n)2+⋯+(21/n)n−1)\begin{align*}\int_0^1 2^x\dee{x}&=\lim_{n \to \infty}\sum_{i=1}^n \Delta x f({a+i\Delta x})\\ &=\lim_{n \to \infty}\sum_{i=1}^n \frac{1}{n} f\left(\frac{i}{n}\right)\\ &=\lim_{n \to \infty}\sum_{i=1}^n \frac{1}{n}\cdot 2^{i/n}\\ &=\lim_{n \to \infty}\frac{1}{n}\left(2^{1/n}+2^{2/n}+2^{3/n}+\cdots + 2^{n/n}\right)\\ &=\lim_{n \to \infty}\frac{2^{1/n}}{n}\left(1+2^{1/n}+2^{2/n}+\cdots + 2^{\frac{n-1}{n}}\right)\\ &=\lim_{n \to \infty}\frac{2^{1/n}}{n}\left(1+2^{1/n}+\left(2^{1/n}\right)^2+\cdots + \left(2^{1/n}\right)^{n-1}\right)\end{align*}

The sum in parenthesis has the form of a geometric sum, with r=21/nr=2^{1/n}:

=lim⁡n→∞21/nn((21/n)n−121/n−1)=lim⁡n→∞21/nn(2−121/n−1)=lim⁡n→∞21/nn(21/n−1)\begin{align*}&=\lim_{n \to \infty}\frac{2^{1/n}}{n}\left( \frac{\left(2^{1/n}\right)^n-1}{2^{1/n}-1} \right)\\ &=\lim_{n \to \infty}\frac{2^{1/n}}{n}\left( \frac{2-1}{2^{1/n}-1} \right)\\ &=\lim_{n \to \infty} \frac{2^{1/n}}{n(2^{1/n}-1)}\end{align*}

Note as n→∞n \to \infty, 1/n→01/n \to 0, so the numerator has limit 1, while the denominator has indeterminate form ∞⋅0\infty\cdot 0. So, we'll do a little algebra to get this into a l'H^opital-style indeterminate form:

=lim⁡n→∞1n⋅21/n21/n−1=lim⁡n→∞1n1−2−1/n⏟num→0den→0\begin{align*}&=\lim_{n \to \infty} \frac{\frac{1}{n}\cdot2^{1/n}}{2^{1/n}-1} \\&=\lim_{n \to \infty} \underbrace{\frac{\frac{1}{n}}{1-2^{-1/n}}}_{\atp{\mathrm{num}\to 0}{\mathrm{den}\to 0}}\end{align*}

Now we can use l'H^opital's rule. Recall ddx{2x}=2xlog⁡x\diff{}{x}\left\{2^x\right\}=2^x\log x, where log⁡x\log x is the natural logarithm of xx, also sometimes written ln⁡x\ln x. We'll need to use the chain rule when we differentiate the denominator.

=lim⁡n→∞−1n2−2−1/nlog⁡2⋅1n2=lim⁡n→∞21/nlog⁡2=1log⁡2\begin{align*}&=\lim_{n \to \infty} \frac{\frac{-1}{n^2}}{-2^{-1/n}\log 2 \cdot \frac{1}{n^2}} \\&=\lim_{n \to \infty} \frac{2^{1/n}}{\log 2}\\ &=\frac{1}{\log 2}\end{align*}

Using a calculator, we see this is about 1.44 square units.

Q44Stage 3

  1. Using the method of Example 1.1.2 in the CLP-2 text, evaluate

    ∫ab10x dx\int_a^b 10^x \ \dee{x}

Using your answer from above, make a guess for

∫abcx dx\int_a^b c^x \ \dee{x}

where cc is a positive constant. Does this agree with Question 43?

Hint

Part (a) follows the same pattern as Question 43–there's just a little more algebra involved, since our lower limit of integration is not 0.

Answer

(a) 1log⁡10(10b−10a)\dfrac{1}{\log 10}\left(10^b-10^a\right)
(b) 1log⁡c(cb−ca)\dfrac{1}{\log c}\left(c^b-c^a\right); yes, it agrees.

Full solution

As in the text, we'll set up a Riemann sum for the given integral. Right Riemann sums have the simplest form:

∫ab10xdx=lim⁡n→∞∑i=1nΔxf(a+iΔx)=lim⁡n→∞∑i=1nb−anf(a+ib−an)=lim⁡n→∞∑i=1nb−an⋅10a+ib−an=lim⁡n→∞∑i=1nb−an⋅10a⋅(10b−an)i=lim⁡n→∞b−an⋅10a((10b−an)1+(10b−an)2+(10b−an)3+⋯+(10b−an)n)=lim⁡n→∞b−an⋅10a⋅10b−an(1+(10b−an)+(10b−an)2+⋯+(10b−an)n−1)\begin{align*}\int_a^b 10^x\dee{x}&=\lim_{n \to \infty}\sum_{i=1}^n \Delta x f({a+i\Delta x})\\ &=\lim_{n \to \infty}\sum_{i=1}^n \frac{b-a}{n} f\left(a+i\frac{b-a}{n}\right)\\ &=\lim_{n \to \infty}\sum_{i=1}^n \frac{b-a}{n}\cdot 10^{a+i\frac{b-a}{n}}\\ &=\lim_{n \to \infty}\sum_{i=1}^n \frac{b-a}{n}\cdot 10^a\cdot \left(10^{\frac{b-a}{n}}\right)^{i}\\ &=\lim_{n \to \infty} \frac{b-a}{n}\cdot 10^a\left(\left(10^{\frac{b-a}{n}}\right)^1+ \left(10^{\frac{b-a}{n}}\right)^2 +\left(10^{\frac{b-a}{n}}\right)^3 +\cdots + \left(10^{\frac{b-a}{n}}\right)^n \right)\\ &=\lim_{n \to \infty} \frac{b-a}{n} \cdot10^{a}\cdot10^{\frac{b-a}{n}} \left(1+\left(10^{\frac{b-a}{n}}\right)+ \left(10^{\frac{b-a}{n}}\right)^2 +\cdots + \left(10^{\frac{b-a}{n}}\right)^{n-1} \right)\end{align*}

Now the sum in parentheses has the form of a geometric sum, with r=10b−anr=10^{\frac{b-a}{n}}:

=lim⁡n→∞b−an⋅10a⋅10b−an((10b−an)n−110b−an−1)=lim⁡n→∞b−an⋅10a⋅10b−an(10b−a−110b−an−1)\begin{align*}&=\lim_{n \to \infty} \frac{b-a}{n}\cdot 10^{a}\cdot 10^{\frac{b-a}{n}}\left( \frac{\left(10^{\frac{b-a}{n}}\right)^n-1}{10^{\frac{b-a}{n}}-1} \right)\\ &=\lim_{n \to \infty} \frac{\textcolor{blue}{b-a}}{n}\cdot \textcolor{red}{10^{a}}\cdot 10^{\frac{b-a}{n}}\left( \frac{\textcolor{purple}{10^{b-a}-1}}{10^{\frac{b-a}{n}}-1} \right)\end{align*}

The coloured parts do not depend on nn, so for simplicity we can move them outside the limit.

=(b−a)⋅10a(10b−a−1)lim⁡n→∞1n⋅(10b−an10b−an−1)=(b−a)⋅(10b−10a)lim⁡n→∞(1/n1−10−b−an)⏟num→0den→0\begin{align*}&=\textcolor{blue}{(b-a)}\cdot\textcolor{red}{10^a}\left(\textcolor{purple}{10^{b-a}-1}\right)\lim_{n \to \infty} \frac{1}{n}\cdot \left( \frac{ 10^{\frac{b-a}{n}} }{10^{\frac{b-a}{n}}-1} \right) \\&={(b-a)}\cdot\left({10^{b}-10^a}\right)\lim_{n \to \infty} \underbrace{\left( \frac{ 1/n}{1-10^{-\frac{b-a}{n}}} \right)}_{\atp{\mathrm{num}\to 0}{\mathrm{den}\to 0}}\end{align*}

Now we can use l'H^opital's rule. Recall ddx{10x}=10xlog⁡x\diff{}{x}\left\{10^x\right\}=10^x\log x, where log⁡x\log x is the natural logarithm of xx, also sometimes written ln⁡x\ln x. For the denominator, we will have to use the chain rule.

=(b−a)⋅(10b−10a)lim⁡n→∞(−1/n2−10−b−an⋅log⁡10⋅b−an2)=(b−a)⋅(10b−10a)lim⁡n→∞(110−b−an⋅log⁡10⋅(b−a))=(b−a)⋅(10b−10a)(1log⁡10⋅(b−a))=1log⁡10(10b−10a)\begin{align*}&={(b-a)}\cdot\left({10^{b}-10^a}\right)\lim_{n \to \infty} \left( \frac{ -1/n^2}{-10^{-\frac{b-a}{n}}\cdot \log 10 \cdot \frac{b-a}{n^2}} \right)\\ &={(b-a)}\cdot\left({10^{b}-10^a}\right)\lim_{n \to \infty} \left( \frac{ 1}{10^{-\frac{b-a}{n}}\cdot \log 10 \cdot (b-a)} \right)\\ &={(b-a)}\cdot\left({10^{b}-10^a}\right) \left( \frac{ 1}{ \log 10 \cdot (b-a)} \right)\\ &=\frac{1}{\log 10}\left(10^b-10^a\right)\end{align*}

For part (b), we can guess that if 10 were changed to cc, our answer would be

∫abcx dx=1log⁡c(cb−ca)\int_a^b c^x \ \dee{x}=\frac{1}{\log c}\left(c^b-c^a\right)

In Question 43, we had a=0a=0, b=1b=1, and c=2c=2. In this case, the formula we guessed above gives

∫012x dx=1log⁡2(21−20)=1log⁡2\int_0^1 2^x \ \dee{x}=\frac{1}{\log 2}\left(2^1-2^0\right)=\frac{1}{\log2}

This does indeed match the answer we calculated.

(In fact, we can directly show ∫abcx dx=1log⁡c(cb−ca)\displaystyle\int_a^b c^x \ \dee{x}=\dfrac{1}{\log c}\left(c^b-c^a\right) using the method of this problem.)

Q45Stage 3

Evaluate ∫0a1−x2 dx\displaystyle\int_0^a \sqrt{1-x^2}\ \dee{x} using geometry, if 0≤a≤10 \leq a \leq 1.

Hint

Your area can be divided into a section of a circle and a triangle. Then you can use geometry to find the area of each piece.

Answer

π4−12arccos⁡(a)+12a1−a2\frac{\pi}{4} -\frac{1}{2} \arccos(a) + \frac{1}{2}a\sqrt{1-a^2}

Full solution

First, we note y=1−x2y=\sqrt{1-x^2} is the upper half of a circle of radius 1, centred at the origin. We're taking the area under the curve from 0 to aa, so the area in question is as shown in the picture below.

Figure from prob_s1.1, line 2

Figure from prob_s1.1, line 2

In order to use geometry to find this area, we break it up into two pieces: a sector of a circle, and a triangle, shown below.

Figure from prob_s1.1, line 2

Figure from prob_s1.1, line 2

  • The sector is a portion of a circle with radius 1, with inner angle θ\theta. So, its area is θ2π(area of circle)=θ2π(π)=θ2\frac{\theta}{2\pi}\left(\text{area of circle}\right) = \frac{\theta}{2\pi}\left(\pi\right) = \frac{\theta}{2}.

    Our job now is to find θ\theta in terms of aa. Note π2−θ\frac{\pi}{2}-\theta is the inner angle of the red triangle, which lies in the unit circle. So, cos⁡(π2−θ)=a\cos\left(\frac{\pi}{2}-\theta\right)=a. Then π2−θ=arccos⁡(a)\frac{\pi}{2}-\theta= \arccos(a), and so θ=π2−arccos⁡(a)\theta = \frac{\pi}{2} - \arccos(a).

    Then the area of the sector is π4−12arccos⁡(a)\frac{\pi}{4} - \frac{1}{2}\arccos(a) square units.

  • The triangle has base aa. Its height is the yy-value of the function when x=ax=a, so its height is 1−a2\sqrt{1-a^2}. Then the area of the triangle is 12a1−a2\frac{1}{2}a\sqrt{1-a^2}.

We conclude ∫0a1−x2 dx=π4−12arccos⁡(a)+12a1−a2\displaystyle\int_0^a \sqrt{1-x^2}\ \dee{x} = \frac{\pi}{4} -\frac{1}{2} \arccos(a) + \frac{1}{2}a\sqrt{1-a^2}.

Q46Stage 3

Suppose f(x)f(x) is a positive, decreasing function from x=ax=a to x=bx=b. You give an upper and lower bound on the area under the curve y=f(x)y=f(x) using nn rectangles and a left and right Riemann sum, respectively, as in the picture below.

Figure from prob_s1.1, line 2

Figure from prob_s1.1, line 2

Figure from prob_s1.1, line 17

Figure from prob_s1.1, line 17

  1. What is the difference between the lower bound and the upper bound? (That is, if we subtract the smaller estimate from the larger estimate, what do we get?) Give your answer in terms of ff, aa, bb, and nn.

  2. If you want to approximate the area under the curve to within 0.01 square units using this method, how many rectangles should you use? That is, what should nn be?

Hint

  1. The difference between the upper and lower bounds is the area that is outside of the smaller rectangles but inside the larger rectangles. Drawing both sets of rectangles on one picture might make things clearer. Look for an easy way to compute the area you want.

  2. Use your answer from Part (a). Your answer will depend on ff, aa, and bb.

Answer

  1. [f(b)−f(a)]⋅b−an\left[f(b)-f(a)\right]\cdot\dfrac{b-a}{n}

  2. Choose nn to be an integer that is greater than or equal to 100[f(b)−f(a)](b−a)100\left[f(b)-f(a)\right](b-a).

Full solution

  1. The difference between our upper and lower bounds is the difference in areas between the larger set of rectangles and the smaller set of rectangles. Drawing them on a single picture makes this a little clearer.

    Figure from prob_s1.1, line 2

    Figure from prob_s1.1, line 2

    Each of the rectangles has width b−an\frac{b-a}{n}, since we took a segment of the xx-axis with length b−ab-a and chopped it into nn pieces. We could calculate the height of each rectangle, but it would be a little complicated, since it differs for each of them. An easier method is to notice that the area we want to calculate can be imagined as a single rectangle:

    Figure from prob_s1.1, line 2

    Figure from prob_s1.1, line 2

    The rectangle has base b−an\frac{b-a}{n}. Its highest coordinate is f(a)f(a), and its lowest is f(b)f(b), so its height is f(b)−f(a)f(b)-f(a). Therefore, the difference in area between our lower bound and our upper bound is:

    [f(b)−f(a)]⋅b−an\left[f(b)-f(a)\right]\cdot\frac{b-a}{n}

  2. We want to give a range with length at most 0.01, and guarantee that the area under the curve y=f(x)y=f(x) is inside that range. In the previous part, we figured out that when we use nn rectangles, the length of our range is [f(b)−f(a)]⋅b−an\left[f(b)-f(a)\right]\cdot\frac{b-a}{n}. So, all we have to do is set this to be less than or equal to 0.01, and solve for nn:

    [f(b)−f(a)]⋅b−an≤0.01100[f(b)−f(a)]⋅(b−a)≤n\begin{align*} \left[f(b)-f(a)\right]\cdot\frac{b-a}{n}&\leq 0.01 \\ 100\left[f(b)-f(a)\right]\cdot(b-a)&\leq n \end{align*}

    We can choose nn to be an integer that is greater than or equal to 100[f(b)−f(a)]⋅(b−a)100\left[f(b)-f(a)\right]\cdot(b-a). Using that many rectangles, we find an upper and lower bound for the area under the curve. If we choose any number between our upper and lower bound as an approximation for the area under the curve, our error is no more than 0.01.

Remark: this question depends on the fact that ff is decreasing and positive from aa to bb. In general, bounding errors on approximations like this is not so straightforward.

Q47Stage 3

Let f(x)f(x) be a linear function, let a<ba<b be integers, and let nn be a whole number. True or false: if we average the left and right Riemann sums for ∫abf(x) dx\displaystyle\int_a^b f(x)\ \dee{x} using nn rectangles, we get the same value as the midpoint Riemann sum using nn rectangles.

Hint

Since f(x)f(x) is linear, there exist real numbers mm and cc such that f(x)=mx+cf(x)=mx+c. It's a little easier to first look at a single triangle from each sum, rather than the sums in their entirety.

Answer

true (but note, for a non-linear function, it is possible that the midpoint Riemann sum is not the average of the other two)

Full solution

Since f(x)f(x) is linear, there exist real numbers mm and cc such that f(x)=mx+cf(x)=mx+c. Now we can do some calculations. Suppose we have a rectangle in our Riemann sum that takes up the interval [x,x+w][x,x+w].

  • If we are using a left Riemann sum, our rectangle has height f(x)=mx+cf(x)=mx+c. Then it has area w(mx+c)w(mx+c).

  • If we are using a right Riemann sum, our rectangle has height f(x+w)=m(x+w)+c=mx+c+mwf(x+w)=m(x+w)+c=mx+c+mw. Then it has area w(mx+c+mw)w(mx+c+mw).

  • If we are using a midpoint Riemann sum, our rectangle has height f(x+12w)=m(x+12w)+c=mx+c+12mwf(x+\frac{1}{2}w)=m(x+\frac{1}{2}w)+c=mx+c+\frac{1}{2}mw. Then it has area w(mx+c+12w)w\left(mx+c+\frac{1}{2}w\right).

So, for each rectangle in our sums, the midpoint rectangle has the same area as the average of the left and right rectangles:

w(mx+c+12mw)=w(mx+c)+w(mx+c+mw)2w\left(mx+c+\frac{1}{2}mw\right) = \dfrac{\textcolor{blue}{w(mx+c)}+\textcolor{red}{w(mx+c+mw)}}{2}

It follows that the midpoint Riemann sum has a value equal to the average of the values of the left and right Riemann sums. To see this, let the rectangles in the midpoint Riemann sum have areas M1,M2,…,MnM_1,M_2,\ldots,M_n, let the rectangles in the left Riemann sum have areas L1,L2,…,Ln\textcolor{blue}{L_1,L_2,\ldots,L_n}, and let the rectangles in the right Riemann sum have areas R1,R2,…,Rn\textcolor{red}{R_1,R_2,\ldots,R_n}. Then the midpoint Riemann sum evaluates to M1+M2+⋯+MnM_1+M_2+\cdots+M_n, and:

[L1+L2+…+Ln]+[R1+R2+…+Rn]2=L1+R12+L2+R22+⋯+Ln+Rn2=M1+M2+⋯+Mn\begin{align*}\dfrac{\textcolor{blue}{[L_1+L_2+\ldots+L_n]}+ \textcolor{red}{[R_1+R_2+\ldots+R_n]} }{2} &= \dfrac{\textcolor{blue}{L_1}+ \textcolor{red}{R_1} }{2}+ \dfrac{\textcolor{blue}{L_2}+ \textcolor{red}{R_2} }{2}+ \cdots + \dfrac{\textcolor{blue}{L_n}+ \textcolor{red}{R_n} }{2} \\ &=M_1+M_2+\cdots+M_n\end{align*}

So, the statement is true.

(Note, however, it is false for many non-linear functions f(x)f(x).)

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From the UBC Math 101 open textbook project, © Joel Feldman, Andrew Rechnitzer, Elyse Yeager and others. Licensed CC BY-NC-SA 4.0; this HTML adaptation is released under the same licence. Full attribution.