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Integration

1.2 Basic properties of the definite integral

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

1Stage 1Conceptual

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

Q1Stage 1

For each of the following properties of definite integrals, draw a picture illustrating the concept, interpreting definite integrals as areas under a curve.

For simplicity, you may assume that a≤c≤ba \leq c \leq b, and that f(x),g(x)f(x),g(x) give positive values.

  1. ∫aaf(x) dx=0\displaystyle\int_a^a f(x)\,\dee{x}=0 (Theorem 1.2.3.a in the CLP-2 text)

  2. ∫abf(x) dx=∫acf(x) dx+∫cbf(x)dx\displaystyle\int_a^b f(x)\,\dee{x}= \displaystyle\int_a^c f(x)\,\dee{x} + \int_c^b f(x)\dee{x} (Theorem 1.2.3.c in the CLP-2 text)

  3. ∫ab(f(x)+g(x)) dx=∫abf(x) dx+∫abg(x) dx\displaystyle\int_a^b \left( f(x) + g(x) \right)\,\dee{x} = \displaystyle\int_a^b f(x)\,\dee{x} + \displaystyle\int_a^b g(x)\,\dee{x} (Theorem 1.2.1.a in the CLP-2 text)

Hint

  1. What is the length of this figure?

  2. Think about cutting the area into two pieces vertically.

  3. Think about cutting the area into two pieces another way.

Answer

Possible drawings:

Figure from prob_s1.2, line 2

Figure from prob_s1.2, line 2

Figure from prob_s1.2, line 9

Figure from prob_s1.2, line 9

Figure from prob_s1.2, line 21

Figure from prob_s1.2, line 21

Full solution

  1. ∫aaf(x) dx=0\displaystyle\int_a^a f(x)\,\dee{x}=0

    Figure from prob_s1.2, line 2

    Figure from prob_s1.2, line 2

    The area under the curve is zero, because it's a region with no width.

  2. ∫abf(x) dx=∫acf(x) dx+∫cbf(x)dx\displaystyle\int_a^b f(x)\,\dee{x}=\textcolor{blue}{ \displaystyle\int_a^c f(x)\,\dee{x}} +\textcolor{red}{ \int_c^b f(x)\dee{x} }

    Figure from prob_s1.2, line 2

    Figure from prob_s1.2, line 2

    If we assume a≤c≤ba \leq c \leq b, then this identity simply tells us that if we add up the area under the curve from aa to cc, and from cc to bb, then we get the whole area under the curve from aa to bb.

    (The situation is slightly more complicated when cc is not between aa and bb, but it still works out.)

  3. ∫ab(f(x)+g(x)) dx=∫abf(x) dx+∫abg(x) dx\displaystyle\int_a^b \left( f(x) + g(x) \right)\,\dee{x} =\textcolor{blue}{ \displaystyle\int_a^b f(x)\,\dee{x}} +\textcolor{red}{ \displaystyle\int_a^b g(x)\,\dee{x}}

    Figure from prob_s1.2, line 21

    Figure from prob_s1.2, line 21

    The blue-shaded area in the picture above is ∫abf(x) dx\displaystyle\int_a^b f(x)\ \dee{x}. The area under the curve f(x)+g(x)f(x)+g(x) but above the curve f(x)f(x) (shown in red) is ∫abg(x) dx\displaystyle\int_a^b g(x)\ \dee{x}.

Q2Stage 1

If ∫0bcos⁡x dx=sin⁡b\displaystyle\int_0^b \cos x\ \dee{x}=\sin b, then what is ∫abcos⁡x dx\displaystyle\int_a^b \cos x\ \dee{x}?

Hint

Use the identity ∫abf(x) dx=∫acf(x) dx+∫cbf(x) dx\int\limits_a^b f(x)\ \dee{x} = \int\limits_a^c f(x)\ \dee{x}+ \int\limits_c^b f(x)\ \dee{x}.

Answer

sin⁡b−sin⁡a\sin b-\sin a

Full solution

Using the identity

∫abf(x) dx=∫acf(x) dx+∫cbf(x) dx ,\begin{align*}\int\limits_a^b f(x)\ \dee{x} &= \int\limits_a^c f(x)\ \dee{x}+ \int\limits_c^b f(x)\ \dee{x}\ ,\end{align*}

we see

∫abcos⁡x dx=∫a0cos⁡x dx+∫0bcos⁡x dx=−∫0acos⁡x dx+∫0bcos⁡x dx=−sin⁡a+sin⁡b=sin⁡b−sin⁡a\begin{align*}\int\limits_a^b\cos x\ \dee{x} &= \int\limits_a^0 \cos x\ \dee{x}+ \int\limits_0^b \cos x\ \dee{x}\\&= -\int\limits_0^a \cos x\ \dee{x}+ \int\limits_0^b \cos x\ \dee{x}\\ &=-\sin a + \sin b\\ &=\sin b - \sin a\end{align*}
Q3Stage 1Past exam · 2015A, 2016A

Decide whether each of the following statements is true or false. If false, provide a counterexample. If true, provide a brief justification.

  1. ∫−3−2f(x)dx=−∫32f(x)dx\displaystyle\int_{-3}^{-2} f(x) \dee{x}=-\displaystyle\int_{3}^{2} f(x) \dee{x}.

  2. If f(x)f(x) is an odd function, then ∫−3−2f(x) dx=∫23f(x) dx\displaystyle \int_{-3}^{-2} f(x)\,\dee{x} = \int_2^3 f(x)\,\dee{x}.

  3. ∫01f(x)⋅g(x) dx=∫01f(x) dx⋅∫01g(x) dx\displaystyle\int_{0}^{1} f(x)\cdot g(x) ~\dee{x} =\int_{0}^{1} f(x) ~\dee{x} \cdot \int_{0}^{1} g(x) ~\dee{x}.

Answer

(a) False. For example, the function

f(x)={0for x<01for x≥0\begin{align*} f(x) = \begin{cases} 0 & \text{for }x<0 \\ 1 & \text{for }x\ge0 \end{cases} \end{align*}

provides a counterexample.

(b) False. For example, the function f(x)=xf(x)=x provides a counterexample.

(c) False. For example, the functions

f(x)={0for x<121for x≥12andg(x)={0for x≥121for x<12\begin{align*} f(x) = \begin{cases} 0 & \text{for }x<\frac{1}{2} \\ 1 & \text{for }x\ge\frac{1}{2} \end{cases} &&\text{and}&&g(x) = \begin{cases} 0 & \text{for }x\ge \frac{1}{2} \\ 1 & \text{for }x<\frac{1}{2} \end{cases} \end{align*}

provide a counterexample.

Full solution

(a) False. For example if

f(x)={0for x<01for x≥0\begin{align*} f(x) = \begin{cases} 0 & \text{for }x<0 \\ 1 & \text{for }x\ge0 \end{cases} \end{align*}

then ∫−3−2f(x)dx=0\int_{-3}^{-2} f(x) \dee{x}=0 and −∫32f(x)dx=∫23f(x)dx=1-\int_{3}^{2} f(x) \dee{x}=\int^{3}_{2} f(x) \dee{x}=1.

Figure from prob_s1.2, line 2

Figure from prob_s1.2, line 2

(b) False. For example, if f(x)=xf(x)=x, then ∫−3−2f(x) dx\int_{-3}^{-2} f(x)\,\dee{x} is negative while ∫23f(x) dx\int_2^3 f(x)\,\dee{x} is positive, so they cannot be the same.

Figure from prob_s1.2, line 2

Figure from prob_s1.2, line 2

(c) False. For example, consider the functions

f(x)={0for x<121for x≥12andg(x)={0for x≥121for x<12\begin{align*} f(x) = \begin{cases} 0 & \text{for }x<\frac{1}{2} \\ 1 & \text{for }x\ge\frac{1}{2} \end{cases} &&\text{and}&&g(x) = \begin{cases} 0 & \text{for }x\ge \frac{1}{2} \\ 1 & \text{for }x<\frac{1}{2} \end{cases} \end{align*}

Then f(x)⋅g(x)=0f(x)\cdot g(x)=0 for all xx, so ∫01f(x)⋅g(x)dx=0\int_0^1 f(x)\cdot g(x) \dee{x}=0. However, ∫01f(x)dx=12\int_0^1 f(x) \dee{x}= \frac{1}{2} and ∫01g(x)dx=12\int_0^1 g(x) \dee{x}= \frac{1}{2}, so ∫01f(x)dx⋅∫01g(x)dx=14\int_0^1 f(x)\dee{x} \cdot \int_0^1 g(x) \dee{x}= \frac{1}{4}.

Figure from prob_s1.2, line 2

Figure from prob_s1.2, line 2

Figure from prob_s1.2, line 12

Figure from prob_s1.2, line 12

Q4Stage 1

Suppose we want to make a right Riemann sum with 100 intervals to approximate ∫50f(x) dx\int\limits_5^0 f(x)\ \dee{x}, where f(x)f(x) is a function that gives only positive values.

  1. What is Δx\Delta x?

  2. Are the heights of our rectangles positive or negative?

  3. Is our Riemann sum positive or negative?

  4. Is the signed area under the curve y=f(x)y=f(x) from x=0x=0 to x=5x=5 positive or negative?

Hint

Note that the limits of the integral given are in the opposite order from what we might expect: the smaller number is the top limit of integration.

Recall Δx=b−an\De x = \frac{b-a}{n}.

Answer

(a) −120-\dfrac{1}{20} (b) positive (c) negative (d) positive

Full solution

  1. Δx=b−an=0−5100=−120\Delta x = \dfrac{b-a}{n}=\dfrac{0-5}{100} = -\dfrac{1}{20}

    Note: if we were to use the Riemann-sum definition of a definite integral, this is how we would justify the identity ∫abf(x)dx=−∫baf(x)dx\int\limits_a^b f(x)\dee{x}=-\int\limits_b^a f(x)\dee{x}.

  2. The heights of the rectangles are given by f(xi)f(x_i), where xi=a+iΔx=5−i20x_i = a+i\Delta x = 5 - \frac{i}{20}. Since f(x)f(x) only gives positive values, f(xi)>0f(x_i) >0, so the heights of the rectangles are positive.

  3. Our Riemann sum is the sum of the signed areas of individual rectangles. Each rectangle has a negative base (Δx\Delta x) and a positive height (f(xi)f(x_i)). So, each term of our sum is negative. If we add up negative numbers, the sum is negative. So, the Riemann sum is negative.

  4. Since f(x)f(x) is always above the xx-axis, ∫05f(x)dx\int\limits_0^5 f(x)\dee{x} is positive.

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q5Stage 2Past exam · M105 2015A

Suppose ∫23f(x) dx=−1\displaystyle\int_2^3 f(x)\,\dee{x} = -1 and ∫23g(x) dx=5\displaystyle\int_2^3 g(x)\,\dee{x} = 5. Evaluate ∫23(6f(x)−3g(x)) dx\displaystyle \int_2^3 \big( 6 f(x) - 3 g(x) \big)\,\dee{x}.

Hint

Split the “target integral” up into pieces that can be evaluated using the given integrals.

Answer

−21-21

Full solution

The operation of integration is linear (that's part (d) of the “arithmetic of integration” Theorem 1.2.1 in the CLP-2 text), so that:

∫23[6f(x)−3g(x)] dx=∫236f(x) dx−∫233g(x) dx=6∫23f(x) dx−3∫23g(x) dx=(6×(−1))−(3×5)=−21\begin{align*} \int_2^3 [6 f(x) -3 g(x)]\,\dee{x} &= \int_2^3 6 f(x)\,\dee{x} - \int_2^3 3 g(x)\,\dee{x} \\ &= 6 \int_2^3 f(x)\,\dee{x} - 3\int_2^3 g(x)\,\dee{x} = (6 \times (-1)) - (3 \times 5) = -21 \end{align*}
Q6Stage 2Past exam · 2016Q1

If ∫02f(x) dx=3\displaystyle\int_0^2 f(x)\,\dee{x} = 3 and ∫02g(x) dx=−4\displaystyle\int_0^2 g(x)\,\dee{x} = -4, calculate ∫02(2f(x)+3g(x)) dx\displaystyle \int_0^2 \big( 2 f(x) + 3 g(x) \big)\,\dee{x}.

Hint

Split the “target integral” up into pieces that can be evaluated using the given integrals.

Answer

−6-6

Full solution

The operation of integration is linear (that's part (d) of the “arithmetic of integration” Theorem 1.2.1 in the CLP-2 text), so that:

∫02[2f(x)+3g(x)] dx=∫022f(x) dx+∫023g(x) dx=2∫02f(x) dx+3∫02g(x) dx=(2×3)+(3×(−4))=−6\begin{align*} \int_0^2 [2 f(x) +3 g(x)]\,\dee{x} &= \int_0^2 2 f(x)\,\dee{x} + \int_0^2 3 g(x)\,\dee{x} \\ &= 2 \int_0^2 f(x)\,\dee{x} + 3\int_0^2 g(x)\,\dee{x} = (2 \times 3) + (3 \times (-4)) = -6 \end{align*}
Q7Stage 2Past exam · 2016Q1

The functions f(x)f(x) and g(x)g(x) obey

∫0−1f(x) dx=1∫02f(x) dx=2∫−10g(x) dx=3∫02g(x) dx=4\begin{equation*} \int_0^{-1} f(x)\,\dee{x} = 1 \qquad \int_0^2 f(x)\,\dee{x} = 2 \qquad \int_{-1}^0 g(x)\,\dee{x} = 3 \qquad \int_0^2 g(x)\,\dee{x} = 4 \end{equation*}

Find ∫−12[3g(x)−f(x)] dx\int_{-1}^2 \big[3g(x)-f(x)\big]\,\dee{x}.

Hint

Split the “target integral” up into pieces that can be evaluated using the given integrals.

Answer

20

Full solution

Using part (d) of the “arithmetic of integration” Theorem 1.2.1, followed by parts (c) and (b) of the “arithmetic for the domain of integration” Theorem 1.2.3 in the

in the CLP-2 text,

∫−12[3g(x)−f(x)] dx=3∫−12g(x) dx−∫−12f(x) dx=3∫−10g(x) dx+3∫02g(x) dx−∫−10f(x) dx−∫02f(x) dx=3∫−10g(x) dx+3∫02g(x) dx+∫0−1f(x) dx−∫02f(x) dx=3×3+3×4+1−2=20\begin{align*} \int_{-1}^2 \big[3g(x)-f(x)\big]\,\dee{x} &=3\int_{-1}^2 g(x)\,\dee{x}-\int_{-1}^2 f(x)\,\dee{x} \\ &=3\int_{-1}^0 g(x)\,\dee{x}+3\int_0^2 g(x)\,\dee{x} -\int_{-1}^0 f(x)\,\dee{x}-\int_0^2 f(x)\,\dee{x} \\ &=3\int^0_{-1} g(x)\,\dee{x}+3\int_0^2 g(x)\,\dee{x} +\int_0^{-1} f(x)\,\dee{x}-\int_0^2 f(x)\,\dee{x} \\ &=3\times 3+3\times 4 + 1 - 2 = 20 \end{align*}
Q8Stage 2

In Question 45, Section 1.1, we found that

∫0a1−x2 dx=π4−12arccos⁡(a)+12a1−a2\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}

when 0≤a≤10\le a\le 1.

Using this fact, evaluate the following:

  1. ∫a01−x2 dx\displaystyle\int_{a}^0 \sqrt{1-x^2}\ \dee{x}, where −1≤a≤0-1 \leq a \leq 0

  2. ∫a11−x2 dx\displaystyle\int_{a}^1 \sqrt{1-x^2}\ \dee{x}, where 0≤a≤10 \leq a \leq 1

Hint

For part (a), use the symmetry of the integrand. For part (b), the area ∫011−x2 dx\int \limits_{0}^1 \sqrt{1-x^2}\ \dee{x} is easy to find–how is this useful to you?

Answer

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

  2. 12arccos⁡(a)−12a1−a2\frac{1}{2}\arccos(a)-\frac{1}{2}a\sqrt{1-a^2}

Full solution

  1. Since 1−x2\sqrt{1-x^2} is an even function,

    ∫a01−x2 dx=∫0∣a∣1−x2 dx=π4−12arccos⁡(∣a∣)+12∣a∣1−∣a∣2=π4−12arccos⁡(−a)−12a1−a2\begin{align*} \int_{a}^0 \sqrt{1-x^2}\ \dee{x} &=\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}\\ &=\frac{\pi}{4} - \frac{1}{2}\arccos(-a)-\frac{1}{2}a\sqrt{1-a^2} \end{align*}

    Alternatively, since arccos⁡(−a)=π−arccos⁡(a)\arccos(-a) = \pi-\arccos(a) we also have

    ∫a01−x2 dx=−π4+12arccos⁡(a)−12a1−a2\begin{align*} \int_{a}^0 \sqrt{1-x^2}\ \dee{x} &=-\frac{\pi}{4} + \frac{1}{2}\arccos(a)-\frac{1}{2}a\sqrt{1-a^2} \end{align*}

  2. Note ∫011−x2 dx=π4\displaystyle\int_{0}^1 \sqrt{1-x^2}\ \dee{x}=\frac{\pi}{4}, since the area under the curve represents one-quarter of the unit circle. Then,

    ∫a11−x2 dx=∫011−x2 dx−∫0a1−x2 dx=π4−(π4−12arccos⁡(a)+12a1−a2)=12arccos⁡(a)−12a1−a2\begin{align*}\displaystyle\int_{a}^1 \sqrt{1-x^2}\ \dee{x}&= \displaystyle\int_{0}^1 \sqrt{1-x^2}\ \dee{x}- \displaystyle\int_{0}^a \sqrt{1-x^2}\ \dee{x}\\ &=\frac{\pi}{4}-\left(\frac{\pi}{4} - \frac{1}{2}\arccos(a)+\frac{1}{2}a\sqrt{1-a^2}\right)\\ &=\frac{1}{2}\arccos(a)-\frac{1}{2}a\sqrt{1-a^2} \end{align*}
Q9Stage 2Past exam · M105 2013A

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

You may use the result from Example 1.2.5 in the CLP-2 text that ∫abx dx=b2−a22\int\limits_a^b x\ \dee{x}=\frac{b^2-a^2}{2}.

Hint

The evaluation of this integral was also the subject of Question 9 in Section 1.1. This time try using the method of Example 1.2.6 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*}

Also recall, from Example 1.2.5 in the

CLP-2 text that

∫abx dx=b2−a22\begin{align*} \int_a^b x\ \dee{x}&=\frac{b^2-a^2}{2} \end{align*}

So

∫−12∣2x∣ dx=∫−10∣2x∣ dx+∫02∣2x∣ dx=∫−10(−2x) dx+∫022x dx=−2∫−10x dx+2∫02x dx=−2⋅02−(−1)22+2⋅22−022=1+4=5\begin{align*} \int_{-1}^2 |2x|\ \dee{x} &=\int_{-1}^0 |2x|\ \dee{x}+\int_0^2 |2x|\ \dee{x} =\int_{-1}^0 (-2x)\ \dee{x}+\int_0^2 2x\ \dee{x} \\ &= -2\int_{-1}^0 x\ \dee{x}+2\int_0^2 x\ \dee{x} =-2\cdot\frac{0^2-(-1)^2}{2} +2\cdot\frac{2^2-0^2}{2} \\ &=1+4=5 \end{align*}
Q10Stage 2

Evaluate ∫−55x∣x∣ dx\displaystyle\int_{-5}^5 x|x|\ \dee{x} .

Hint

Use symmetry.

Answer

0

Full solution

We note that the integrand f(x)=x∣x∣f(x)=x|x| is an odd function, because f(−x)=−x∣−x∣=−x∣x∣=−f(x)f(-x)=-x|-x|=-x|x|=-f(x). Then, by Theorem 1.2.11.b in the CLP-2 text, ∫−55x∣x∣ dx=0\displaystyle\int_{-5}^5 x|x| \ \dee{x}=0.

Q11Stage 2

Suppose f(x)f(x) is an even function and ∫−22f(x)dx=10\displaystyle\int_{-2}^2 f(x)\dee{x}=10. What is ∫−20f(x)dx\displaystyle\int_{-2}^0 f(x)\dee{x}?

Hint

Check Theorem 1.2.11 in the CLP-2 text.

Answer

5

Full solution

Using Theorem 1.2.11.a in the CLP-2 text,

10=∫−22f(x)dx=2∫02f(x)dx5=∫02f(x)dx\begin{align*}10&=\int_{-2}^2 f(x)\dee{x}=2\int_{0}^2f(x)\dee{x}\\ 5&=\int_{0}^2f(x)\dee{x}\end{align*}

Also,

∫−22f(x)dx=∫−20f(x)dx+∫02f(x)dx\begin{align*}\int_{-2}^2 f(x)\dee{x}&=\int_{-2}^0 f(x)\dee{x}+\int_{0}^2 f(x)\dee{x}\end{align*}

So,

∫−20f(x)dx=∫−22f(x)dx−∫02f(x)dx=10−5=5\begin{align*}\int_{-2}^0 f(x)\dee{x}&=\int_{-2}^2 f(x)\dee{x}-\int_{0}^2 f(x)\dee{x}\\ &=10-5=5\end{align*}

Indeed, for any even function f(x)f(x), ∫−a0f(x)dx=∫0af(x)dx\int\limits_{-a}^0 f(x)\dee{x} = \int\limits_{0}^a f(x)\dee{x}.

3Stage 3Application

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

Q12Stage 3Past exam · 2016Q1

Evaluate ∫−22(5+4−x2)dx\displaystyle\int_{-2}^{2} \left(5+\sqrt{4-x^2}\right)\dee{x}.

Hint

Split the integral into a sum of two integrals. Interpret each geometrically.

Answer

20+2π20 +2\pi

Full solution

We first use additivity:

∫−22(5+4−x2)dx=∫−225 dx+∫−224−x2 dx\begin{align*} \int_{-2}^{2} \left(5+\sqrt{4-x^2}\right)\dee{x} = \textcolor{blue}{ \int_{-2}^{2} 5\,\dee{x} }+ \textcolor{red}{\int_{-2}^{2} \sqrt{4-x^2}\,\dee{x}} \end{align*}

The first integral represents the area of a rectangle of height 5 and width 4 and so equals 2020. The second integral represents the area above the xx–axis and below the curve y=4−x2y=\sqrt{4-x^2} or x2+y2=4x^2+y^2=4. That is a semicircle of radius 2, which has area 12π22\frac{1}{2}\pi 2^2. So

∫−22(5+4−x2)dx=20+2π\begin{align*} \int_{-2}^{2} \left(5+\sqrt{4-x^2}\right)\dee{x} = \textcolor{blue}{20} +\textcolor{red}{2\pi} \end{align*}

Figure from prob_s1.2, line 2

Figure from prob_s1.2, line 2

Figure from prob_s1.2, line 12

Figure from prob_s1.2, line 12

Q13Stage 3Past exam · M121 2012A

Evaluate ∫−2012+2012sin⁡xlog⁡(3+x2)dx\displaystyle\int_{-2012}^{+2012} \frac{\sin x}{\log(3+x^2)}\dee{x}.

Hint

Hmmmm. Looks like a complicated integral. It's probably a trick question. Check for symmetries.

Answer

00

Full solution

Note that the integrand f(x)=sin⁡xlog⁡(3+x2)f(x) = \frac{\sin x}{\log(3+x^2)} is an odd function, because:

f(−x)=sin⁡(−x)log⁡(3+(−x)2)=−sin⁡xlog⁡(3+x2)=−f(x)\begin{equation*} f(-x) = \frac{\sin(-x)}{\log(3+(-x)^2)}=\frac{-\sin x}{\log(3+x^2)} =- f(x) \end{equation*}

The domain of integration −2012≤x≤2012-2012 \le x \le 2012 is symmetric about x=0x=0. So, by Theorem 1.2.11 of the CLP-2 text,

∫−2012+2012sin⁡xlog⁡(3+x2)dx=0\begin{equation*} \int_{-2012}^{+2012} \frac{\sin x}{\log(3+x^2)}\dee{x} = 0 \end{equation*}
Q14Stage 3Past exam · 2012A

Evaluate ∫−2012+2012x1/3cos⁡x dx\displaystyle\int_{-2012}^{+2012} x^{1/3}\cos x\,\dee{x}.

Hint

Check for symmetries again.

Answer

00

Full solution

Note that the integrand f(x)=x1/3cos⁡xf(x) = x^{1/3}\cos x is an odd function, because:

f(−x)=(−x)1/3cos⁡(−x)=−x1/3cos⁡x=−f(x)\begin{equation*} f(-x) = (-x)^{1/3}\cos(-x)= - x^{1/3}\cos x =- f(x) \end{equation*}

The domain of integration −2012≤x≤2012-2012 \le x \le 2012 is symmetric about x=0x=0. So, by Theorem 1.2.11 of the CLP-2 text,

∫−2012+2012x1/3cos⁡x dx=0\begin{equation*} \int_{-2012}^{+2012}x^{1/3}\cos x\,\dee{x} = 0 \end{equation*}
Q15Stage 3

Evaluate ∫06(x−3)3 dx\displaystyle\int_{0}^6 (x-3)^3\,\dee{x} .

Hint

What does the integrand look like to the left and right of x=3x=3?

Answer

0

Full solution

Our integrand f(x)=(x−3)3f(x)=(x-3)^3 is neither even nor odd. However, it does have a similar symmetry. Namely, f(3+x)=−f(3−x)f(3+x)=-f(3-x). So, ff is “negatively symmetric" across the line x=3x=3. This suggests that the integral should be 0: the positive area to the right of x=3x=3 will be the same as the negative area to the left of x=3x=3.

Another way to see this is to notice that the graph of f(x)=(x−3)3f(x)=(x-3)^3 is equivalent to the graph of g(x)=x3g(x)=x^3 shifted three units to the right, and g(x)g(x) is an odd function. So,

∫06(x−3)3 dx=∫−33x3 dx=0\textcolor{red}{\int_{0}^6 (x-3)^3\,\dee{x}} = \textcolor{blue}{\int_{-3}^3 x^3\,\dee{x}}=0

Figure from prob_s1.2, line 2

Figure from prob_s1.2, line 2

Q16Stage 3

We want to compute the area of an ellipse, (ax)2+(by)2=1(ax)^2+(by)^2=1 for some (let's say positive) constants aa and bb.

  1. Solve the equation for the upper half of the ellipse. It should have the form “y=⋯y=\cdots"

  2. Write an integral for the area of the upper half of the ellipse. Using properties of integrals, make the integrand look like the upper half of a circle.

  3. Using geometry and your answer to part (b), find the area of the ellipse.

Hint

In part (b), you'll have to factor a constant out through a square root. Remember the upper half of a circle looks like r2−x2\sqrt{r^2-x^2}.

Answer

(a) y=1b1−(ax)2y = \dfrac{1}{b}\sqrt{1-(ax)^2} (b) ab∫−1a1a1a2−x2 dx\displaystyle\frac{a}{b}\int_{-\frac{1}{a}}^{\frac{1}{a}}\sqrt{\frac{1}{a^2}-x^2}\ \dee{x} (c) πab\dfrac{\pi}{ab}

Full solution
  1. (ax)2+(by)2=1by=1−(ax)2y=1b1−(ax)2\begin{align*} (ax)^2+(by)^2&=1\\ by&=\sqrt{1-(ax)^2}\\ y &= \frac{1}{b}\sqrt{1-(ax)^2} \end{align*}

  2. The values of xx in the domain of the function above are those that satisfy 1−(ax)2≥01-(ax)^2 \geq 0. That is, −1a≤x≤1a-\frac{1}{a}\leq x \leq \frac{1}{a}. Therefore, the upper half of the ellipse has area

    1b∫−1a1a1−(ax)2 dx\begin{align*}\displaystyle\frac{1}{b}&\displaystyle\int_{-\frac{1}{a}}^{\frac{1}{a}}\sqrt{1-(ax)^2}\ \dee{x}\end{align*}

    The upper half of a circle has equation y=r2−x2y=\sqrt{r^2-x^2}.

    =1b∫−1a1aa2(1a2−x2) dx=1b∫−1a1aa1a2−x2 dx=ab∫−1a1a1a2−x2 dx\begin{align*}&=\frac{1}{b}\int_{-\frac{1}{a}}^{\frac{1}{a}}\sqrt{a^2\left(\frac{1}{a^2}-x^2\right)}\ \dee{x}\\ &=\frac{1}{b}\int_{-\frac{1}{a}}^{\frac{1}{a}}a\sqrt{\frac{1}{a^2}-x^2}\ \dee{x}\\ &=\frac{a}{b}\int_{-\frac{1}{a}}^{\frac{1}{a}}\sqrt{\frac{1}{a^2}-x^2}\ \dee{x}\end{align*}

  3. The function y=1a2−x2y=\sqrt{\dfrac{1}{a^2}-x^2} is the upper-half of the circle centred at the origin with radius 1a\dfrac{1}{a}. So, the expression from (b) evaluates to (ab)π2a2=π2ab\left(\dfrac{a}{b}\right)\dfrac{\pi}{2a^2} = \dfrac{\pi}{2ab}.

    The expression from (b) was half of the ellipse, so the area of the ellipse is πab\dfrac{\pi}{ab}.

Remark: this was a slightly long-winded way of getting the result. The reasoning is basically this:

  • The area of the unit circle x2+y2=1x^2+y^2=1 is π\pi .

  • The ellipse (ax)2+y2=1(ax)^2+y^2=1 is obtained by shrinking the unit circle horizontally by a factor of aa. So, its area is πa\dfrac{\pi}{a} .

  • Further, the ellipse (ax)2+(by)2=1(ax)^2+(by)^2=1 is obtained from the previous ellipse by shrinking it vertically by a factor of bb. So, its area is πab\dfrac{\pi}{ab} .

Q17Stage 3

Fill in the following table: the product of an (even/odd) function with an (even/odd) function is an (even/odd) function. You may assume that both functions are defined for all real numbers.

×\timesevenodd
even
odd
Hint

For two functions f(x)f(x) and g(x)g(x), define h(x)=f(x)⋅g(x)h(x)=f(x)\cdot g(x). If h(−x)=h(x)h(-x)=h(x), then the product is even; if h(−x)=−h(x)h(-x)=-h(x), then the product is odd.

The table will not be the same as if we were multiplying even and odd numbers.

Answer
×\timesevenodd
evenevenodd
oddoddeven
Full solution

Let's recall the definitions of even and odd functions: f(x)f(x) is even if f(−x)=f(x)f(-x)=f(x) for every xx in its domain, and f(x)f(x) is odd if f(−x)=−f(x)f(-x)=-f(x) for every xx in its domain.

Let h(x)=f(x)⋅g(x)h(x)=f(x)\cdot g(x).

  • If ff and gg are both even, then h(−x)=f(−x)⋅g(−x)=f(x)⋅g(x)=h(x)h(-x)=f(-x)\cdot g(-x) = f(x)\cdot g(x)=h(x), so their product is even.

  • If ff and gg are both odd, then h(−x)=f(−x)⋅g(−x)=[−f(x)]⋅[−g(x)]=f(x)⋅g(x)=h(x)h(-x)=f(-x)\cdot g(-x) =[- f(x)]\cdot [-g(x)]=f(x)\cdot g(x)=h(x), so their product is even.

  • If ff is even and gg is odd, then h(−x)=f(−x)⋅g(−x)=f(x)⋅[−g(x)]=−[f(x)⋅g(x)]=−h(x)h(-x)=f(-x)\cdot g(-x) = f(x)\cdot[- g(x)]=-[f(x)\cdot g(x)]=-h(x), so their product is odd. Because multiplication is commutative, the order we multiply the functions in doesn't matter.

We note that the table would be the same as if we were adding (not multiplying) even and odd numbers (not functions).

Q18Stage 3

Suppose f(x)f(x) is an odd function and g(x)g(x) is an even function, both defined at x=0x=0. What are the possible values of f(0)f(0) and g(0)g(0)?

Hint

Note f(0)=f(−0)f(0)=f(-0).

Answer

f(0)=0f(0)=0; g(0)g(0) can be any real number

Full solution

Since f(x)f(x) is odd, f(0)=−f(−0)=−f(0)f(0)=-f(-0)=-f(0). So, f(0)=0f(0)=0.

However, this restriction does not apply to g(x)g(x). For example, for any constant cc, let g(x)=cg(x)=c. Then g(x)g(x) is even and g(0)=cg(0)=c. So, g(0)g(0) can be any real number.

Q19Stage 3

Suppose f(x)f(x) is a function defined on all real numbers that is both even and odd. What could f(x)f(x) be?

Hint

If f(x)f(x) is even and odd, then f(x)=−f(x)f(x)=-f(x) for every xx.

Answer

f(x)=0f(x)=0 for every xx

Full solution

Let xx be any real number.

  • f(x)=f(−x)f(x)=f(-x) (since f(x)f(x) is even), and

  • f(x)=−f(−x)f(x)=-f(-x) (since f(x)f(x) is odd).

  • So, f(x)=−f(x)f(x)=-f(x).

  • Then (adding f(x)f(x) to both sides) we see 2f(x)=02f(x)=0, so f(x)=0f(x)=0.

So, f(x)=0f(x)=0 for every xx.

Q20Stage 3

Is the derivative of an even function even or odd? Is the derivative of an odd function even or odd?

Hint

Think about mirroring a function across an axis. What does this do to the slope?

Answer

The derivative of an even function is odd, and the derivative of an odd function is even.

Full solution

  • Suppose f(x)f(x) is an odd function. We investigate f′(x)f'(x) using the chain rule:

    f(−x)=−f(x)(odd function)ddx{f(−x)}=ddx{−f(x)}−f′(−x)=−f′(x)(chain rule)f′(−x)=f′(x)\begin{alignat*}{3} f(-x)&=-f(x)& \text{(odd function)}\\ \diff{}{x}\{f(-x)\}&=\diff{}{x}\{-f(x)\}\\ -f'(-x)&=-f'(x) & \text{(chain rule)}\\ f'(-x)&=f'(x) \end{alignat*}

    So, when f(x)f(x) is odd, f′(x)f'(x) is even.

    Similarly, suppose f(x)f(x) is even.

    f(−x)=f(x)(even function)ddx{f(−x)}=ddx{f(x)}−f′(−x)=f′(x)(chain rule)f′(−x)=−f′(x)\begin{alignat*}{3} f(-x)&=f(x)& \text{(even function)}\\ \diff{}{x}\{f(-x)\}&=\diff{}{x}\{f(x)\}\\ -f'(-x)&=f'(x) & \text{(chain rule)}\\ f'(-x)&=-f'(x) \end{alignat*}

    So, when f(x)f(x) is even, f′(x)f'(x) is odd.

  • Another way to think about this problem is to notice that “mirroring" a function changes the sign of its derivative. Then since an even function is “mirrored once" (across the yy-axis), it should have f′(x)=−f′(−x)f'(x)=-f'(-x), and so the derivative of an even function should be an odd function. Since an odd function is “mirrored twice" (across the yy-axis and across the xx-axis), it should have f′(x)=−(−f′(−x))=f′(−x)f'(x)=-(-f'(-x))=f'(-x). So the derivative of an odd function should be even. These ideas are presented in more detail below.

    First, we consider the case where f(x)f(x) is even, and investigate f′(x)f'(x).

    Figure from prob_s1.2, line 2

    Figure from prob_s1.2, line 2

    The whole function has a mirror-like symmetry across the yy-axis. So, at xx and −x-x, the function will have the same “steepness," but if one is increasing then the other is decreasing. That is, f′(−x)=−f′(x)f'(-x)=-f'(x). (In the picture above, compare the slope at some point aia_i with its corresponding point −ai-a_i.) So, f′(x)f'(x) is odd when f(x)f(x) is even.

    Second, let's consider the case where f(x)f(x) is odd, and investigate f′(x)f'(x). Suppose the blue graph below is y=f(x)y=f(x). If f(x)f(x) were even, then to the left of the yy-axis, it would look like the orange graph, which we'll call y=g(x)y=g(x).

    Figure from prob_s1.2, line 2

    Figure from prob_s1.2, line 2

    From our work above, we know that, for every x>0x>0, −f′(x)=g′(−x)-f'(x)=g'(-x). When x<0x<0, f(x)=−g(x)f(x)=-g(x). So, if x>0x>0, then −f′(x)=g′(−x)=−f′(−x)-f'(x)=g'(-x)=-f'(-x). In other words, f′(x)=f′(−x)f'(x)=f'(-x). Similarly, if x<0x<0, then f′(x)=−g′(x)=f′(−x)f'(x)=-g'(x)=f'(-x). Therefore f′(x)f'(x) is even. (In the graph below, you can anecdotally verify that f′(ai)=f′(−ai)f'(a_i)=f'(-a_i).)

    Figure from prob_s1.2, line 2

    Figure from prob_s1.2, line 2

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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.