Topic 1

The Definite Integral & Riemann Sums

A definite integral ∫abf(x) dx\displaystyle\int_a^b f(x)\,dx is the signed area between y=f(x)y = f(x) and the xx-axis: area above counts plus, area below counts minus. A Riemann sum estimates it with nn rectangles of width Δx=b−an\Delta x = \dfrac{b-a}{n}. As n→∞n \to \infty, the sum tends to the exact value.

6 kinds of question. Open the one you are stuck on.

1

Estimate an integral with a left, right or midpoint Riemann sum

e.g. “Estimate ∫abf(x) dx\displaystyle\int_a^b f(x)\,dx using nn approximating rectangles and left (right) endpoints”
  1. 1
    Find the width: Δx=b−an\Delta x = \dfrac{b-a}{n}

    From a table, it is the spacing of the readings you use.

  2. 2
    List the sample points for the sum you were asked for

    Left: a,a+Δx,…,b−Δxa, a + \Delta x, \dots, b - \Delta x. Right: a+Δx,…,ba + \Delta x, \dots, b. Midpoint: the middle of each piece. In , xi=a+i Δxx_i = a + i\,\Delta x is the ii-th right endpoint.

  3. 3
    Evaluate ff at each sample point
  4. 4
    Add the heights and multiply by Δx\Delta x

    In sigma notation: ∑i=1nf(xi∗) Δx\displaystyle\sum_{i=1}^{n} f(x_i^*)\,\Delta x.

Remember: Width, sample points, heights, add.

See it in a problem3 worked examples

Estimate ∫131x dx\displaystyle\int_1^3 \frac{1}{x}\,dx using four rectangles and right endpoints.

  1. 1
    Find the width: Δx=b−an\Delta x = \dfrac{b-a}{n}

    Δx=3−14=12\Delta x = \dfrac{3 - 1}{4} = \dfrac{1}{2}.

  2. 2
    List the sample points for the sum you were asked for

    Right endpoints: 1.5, 2, 2.5, 31.5,\ 2,\ 2.5,\ 3.

  3. 3
    Evaluate ff at each sample point

    Heights: 23, 12, 25, 13\dfrac{2}{3},\ \dfrac{1}{2},\ \dfrac{2}{5},\ \dfrac{1}{3}.

  4. 4
    Add the heights and multiply by Δx\Delta x

    12(23+12+25+13)=12⋅1910=1920\dfrac{1}{2}\left(\dfrac{2}{3} + \dfrac{1}{2} + \dfrac{2}{5} + \dfrac{1}{3}\right) = \dfrac{1}{2} \cdot \dfrac{19}{10} = \dfrac{19}{20}.

Answer:

1920=0.95\dfrac{19}{20} = 0.95 (an underestimate, since 1x\dfrac{1}{x} is decreasing)

Common mistakes3
  • Using n+1n + 1 heights. nn rectangles use nn heights: a left sum skips bb, a right sum skips aa.
  • For an increasing ff, the left sum is an underestimate and the right sum an overestimate. For a decreasing ff, it is the other way around.
  • Where f<0f < 0, the rectangle's signed area is negative.
2

Evaluate a sum in sigma notation using summation formulas

e.g. “Evaluate ∑i=115(i2+4i−2)\displaystyle\sum_{i=1}^{15}\left(i^2 + 4i - 2\right)”
  1. 1
    Split into constant multiples of standard sums

    ∑(c ai+bi)=c∑ai+∑bi\sum (c\,a_i + b_i) = c\sum a_i + \sum b_i. A constant cc summed nn times is ncnc.

  2. 2
    Make each sum start where its formula starts

    ∑i=mnai=∑i=1nai−∑i=1m−1ai\displaystyle\sum_{i=m}^{n} a_i = \sum_{i=1}^{n} a_i - \sum_{i=1}^{m-1} a_i. For a , factor out the first term instead.

  3. 3
    Apply the formulas

    ∑i=1ni=n(n+1)2\sum_{i=1}^{n} i = \dfrac{n(n+1)}{2},  ∑i=1ni2=n(n+1)(2n+1)6\ \sum_{i=1}^{n} i^2 = \dfrac{n(n+1)(2n+1)}{6},  ∑i=1ni3=[n(n+1)2]2\ \sum_{i=1}^{n} i^3 = \left[\dfrac{n(n+1)}{2}\right]^2,  ∑i=0nri=1−rn+11−r\ \sum_{i=0}^{n} r^i = \dfrac{1 - r^{n+1}}{1 - r} for r≠1r \neq 1.

  4. 4
    Simplify

Remember: Split, shift, formula.

See it in a problem3 worked examples

Evaluate ∑i=112(2i2−i+3)\displaystyle\sum_{i=1}^{12}\left(2i^2 - i + 3\right).

  1. 1
    Split into constant multiples of standard sums

    2∑i=112i2−∑i=112i+∑i=11232\sum_{i=1}^{12} i^2 - \sum_{i=1}^{12} i + \sum_{i=1}^{12} 3.

  2. 2
    Make each sum start where its formula starts

    All three already start at i=1i = 1.

  3. 3
    Apply the formulas

    2⋅12⋅13⋅256−12⋅132+3⋅12=2(650)−78+362 \cdot \dfrac{12 \cdot 13 \cdot 25}{6} - \dfrac{12 \cdot 13}{2} + 3 \cdot 12 = 2(650) - 78 + 36.

  4. 4
    Simplify

    1300−78+36=12581300 - 78 + 36 = 1258.

Answer:

12581258

Common mistakes3
  • Miscounting terms: ∑i=mn\sum_{i=m}^{n} has n−m+1n - m + 1 terms, and ∑i=0nri\sum_{i=0}^{n} r^i has n+1n + 1.
  • ∑i2≠(∑i)2\sum i^2 \neq \left(\sum i\right)^2 and ∑aibi≠∑ai∑bi\sum a_i b_i \neq \sum a_i \sum b_i.
  • Writing ∑i=1n3=3\sum_{i=1}^{n} 3 = 3. It is 3n3n.
3

Evaluate a definite integral as a limit of Riemann sums

e.g. “Using a limit of right-endpoint Riemann sums, evaluate ∫abf(x) dx\displaystyle\int_a^b f(x)\,dx”
  1. 1
    Write Δx=b−an\Delta x = \dfrac{b-a}{n} and xi=a+i Δxx_i = a + i\,\Delta x
  2. 2
    Write the right sum Rn=∑i=1nf(xi) ΔxR_n = \displaystyle\sum_{i=1}^{n} f(x_i)\,\Delta x
  3. 3
    Expand, and pull everything without ii out of the sum

    nn is a constant inside the sum; only ii changes.

  4. 4
    Apply the summation formulas
  5. 5
    Take the limit as n→∞n \to \infty

    Divide top and bottom by the highest power of nn. Check with an antiderivative if allowed.

Remember: Set up, expand, formulas, limit.

See it in a problem2 worked examples

Using a limit of right-endpoint Riemann sums, evaluate ∫02(3x2+1) dx\displaystyle\int_0^2 \left(3x^2 + 1\right)\,dx.

  1. 1
    Write Δx=b−an\Delta x = \dfrac{b-a}{n} and xi=a+i Δxx_i = a + i\,\Delta x

    Δx=2n\Delta x = \dfrac{2}{n},  xi=2in\ x_i = \dfrac{2i}{n}.

  2. 2
    Write the right sum Rn=∑i=1nf(xi) ΔxR_n = \displaystyle\sum_{i=1}^{n} f(x_i)\,\Delta x

    Rn=∑i=1n(12i2n2+1)2nR_n = \displaystyle\sum_{i=1}^{n}\left(\frac{12i^2}{n^2} + 1\right)\frac{2}{n}.

  3. 3
    Expand, and pull everything without ii out of the sum

    Rn=24n3∑i=1ni2+2n∑i=1n1R_n = \dfrac{24}{n^3}\displaystyle\sum_{i=1}^{n} i^2 + \frac{2}{n}\sum_{i=1}^{n} 1.

  4. 4
    Apply the summation formulas

    Rn=24n3⋅n(n+1)(2n+1)6+2n⋅n=4(n+1)(2n+1)n2+2R_n = \dfrac{24}{n^3} \cdot \dfrac{n(n+1)(2n+1)}{6} + \dfrac{2}{n} \cdot n = \dfrac{4(n+1)(2n+1)}{n^2} + 2.

  5. 5
    Take the limit as n→∞n \to \infty

    lim⁡n→∞Rn=4⋅2+2=10\displaystyle\lim_{n\to\infty} R_n = 4 \cdot 2 + 2 = 10. Check: [x3+x]02=10\left[x^3 + x\right]_0^2 = 10.

Answer:

1010

Common mistakes3
  • Pulling ii out of the sum. Only factors without ii can come out.
  • ∑i=1n1=n\sum_{i=1}^{n} 1 = n, not 11.
  • Forgetting the factor Δx\Delta x. Without it, the sum usually grows without bound.
4

Write a limit of Riemann sums as a definite integral

e.g. “Express lim⁡n→∞∑i=1n…\displaystyle\lim_{n\to\infty}\sum_{i=1}^{n}\dots as a definite integral”
  1. 1
    Find Δx\Delta x: the factor like 4n\frac{4}{n} multiplying each term

    Split a hidden one: in2=in⋅1n\dfrac{i}{n^2} = \dfrac{i}{n}\cdot\dfrac{1}{n}.

  2. 2
    Spot xi=a+i Δxx_i = a + i\,\Delta x inside the function
  3. 3
    Read off aa; then b=a+n Δxb = a + n\,\Delta x
  4. 4
    Replace xix_i by xx and write the integral

Remember: Find the width, find the x, read the ends.

See it in a problem3 worked examples

Express lim⁡n→∞∑i=1n4n1+4in\displaystyle\lim_{n\to\infty}\sum_{i=1}^{n}\frac{4}{n}\sqrt{1 + \frac{4i}{n}} as a definite integral.

  1. 1
    Find Δx\Delta x: the factor like 4n\frac{4}{n} multiplying each term

    Δx=4n\Delta x = \dfrac{4}{n}.

  2. 2
    Spot xi=a+i Δxx_i = a + i\,\Delta x inside the function

    xi=1+4in=1+i Δxx_i = 1 + \dfrac{4i}{n} = 1 + i\,\Delta x.

  3. 3
    Read off aa; then b=a+n Δxb = a + n\,\Delta x

    a=1a = 1,  b=1+4=5\ b = 1 + 4 = 5.

  4. 4
    Replace xix_i by xx and write the integral

    f(xi)=xif(x_i) = \sqrt{x_i}, so f(x)=xf(x) = \sqrt{x}.

Answer:

∫15x dx\displaystyle\int_1^5 \sqrt{x}\,dx (or ∫041+x dx\displaystyle\int_0^4 \sqrt{1 + x}\,dx)

Common mistakes2
  • Mismatched pieces: the width b−ab - a must equal n Δxn\,\Delta x.
  • Thinking there is only one answer. Shifting xx gives other correct integrals, such as ∫041+x dx=∫15x dx\int_0^4 \sqrt{1 + x}\,dx = \int_1^5 \sqrt{x}\,dx.
5

Evaluate a definite integral using geometry and signed area

e.g. “Evaluate the following integral by interpreting it as a signed area, and using geometry”
  1. 1
    Sketch y=f(x)y = f(x) from x=ax = a to x=bx = b

    y=r2−x2y = \sqrt{r^2 - x^2} is the top half of the circle of radius rr centered at the origin.

  2. 2
    Cut the region into triangles, rectangles and circle pieces
  3. 3
    Find each area; below the axis counts negative
  4. 4
    Add the signed areas

Remember: Sketch, cut, sign, add.

See it in a problem3 worked examples

Evaluate ∫−13(2−x) dx\displaystyle\int_{-1}^{3}(2 - x)\,dx using geometry.

  1. 1
    Sketch y=f(x)y = f(x) from x=ax = a to x=bx = b

    A line through (−1,3)(-1, 3), (2,0)(2, 0) and (3,−1)(3, -1).

  2. 2
    Cut the region into triangles, rectangles and circle pieces

    A triangle above the axis on [−1,2][-1, 2] and one below on [2,3][2, 3].

  3. 3
    Find each area; below the axis counts negative

    Above: 12(3)(3)=92\tfrac{1}{2}(3)(3) = \tfrac{9}{2}. Below: −12(1)(1)=−12-\tfrac{1}{2}(1)(1) = -\tfrac{1}{2}.

  4. 4
    Add the signed areas

    92−12=4\tfrac{9}{2} - \tfrac{1}{2} = 4.

Answer:

44

Common mistakes3
  • Counting area below the axis as positive. ∫abf(x) dx\int_a^b f(x)\,dx is signed area, not total area.
  • Using a full circle for r2−x2\sqrt{r^2 - x^2}: it is only the top half.
  • Treating ∣x+1∣|x + 1| as a line. It is a V: split it at x=−1x = -1.
6

Use properties of definite integrals and odd or even symmetry

e.g. “Suppose ∫abf(x) dx=…\displaystyle\int_a^b f(x)\,dx = \dots and ∫abg(x) dx=…\displaystyle\int_a^b g(x)\,dx = \dots. Evaluate …”
  1. 1
    Flip or split the limits until they match

    ∫baf=−∫abf\displaystyle\int_b^a f = -\int_a^b f and ∫abf=∫acf+∫cbf\displaystyle\int_a^b f = \int_a^c f + \int_c^b f.

  2. 2
    Pull out constants and split sums

    ∫(cf+g)=c∫f+∫g\int (c f + g) = c\int f + \int g. A product does not split.

  3. 3
    On [−a,a][-a, a], test f(−x)f(-x) for odd or even

    Odd, f(−x)=−f(x)f(-x) = -f(x): the integral is 00. Even, f(−x)=f(x)f(-x) = f(x): ∫−aaf=2∫0af\displaystyle\int_{-a}^{a} f = 2\int_0^a f.

  4. 4
    Plug in the known values and add

Remember: Match the limits, split, use symmetry.

See it in a problem3 worked examples

Suppose ∫05f(x) dx=4\displaystyle\int_0^5 f(x)\,dx = 4,  ∫02f(x) dx=1\ \int_0^2 f(x)\,dx = 1 and  ∫52g(x) dx=3\ \int_5^2 g(x)\,dx = 3. Evaluate ∫25(2f(x)+g(x)) dx\displaystyle\int_2^5 \big(2f(x) + g(x)\big)\,dx.

  1. 1
    Flip or split the limits until they match

    ∫25f=∫05f−∫02f=3\displaystyle\int_2^5 f = \int_0^5 f - \int_0^2 f = 3, and ∫25g=−∫52g=−3\displaystyle\int_2^5 g = -\int_5^2 g = -3.

  2. 2
    Pull out constants and split sums

    ∫25(2f+g)=2∫25f+∫25g\displaystyle\int_2^5 (2f + g) = 2\int_2^5 f + \int_2^5 g.

  3. 3
    On [−a,a][-a, a], test f(−x)f(-x) for odd or even

    No symmetric limits here; nothing to test.

  4. 4
    Plug in the known values and add

    2(3)+(−3)=32(3) + (-3) = 3.

Answer:

33

Common mistakes3
  • ∫abf(x)g(x) dx≠∫abf(x) dx⋅∫abg(x) dx\displaystyle\int_a^b f(x)g(x)\,dx \neq \int_a^b f(x)\,dx \cdot \int_a^b g(x)\,dx.
  • Calling an integral 00 because it "looks odd": check f(−x)=−f(x)f(-x) = -f(x), and the limits must be −a-a to aa.
  • Odd times odd is even; odd times even is odd.

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