AP Calculus AB · Lesson 2

Area between curves

Know Know it

If f(x)≥g(x)f(x) \ge g(x) on [a,b][a, b], the area between the curves is ∫ab(f(x)−g(x))dx\int_a^b \left(f(x) - g(x)\right)dx: top minus bottom. The limits are usually the x-values where the curves meet, found by setting f(x)=g(x)f(x) = g(x).

If the curves cross inside the interval, the top and bottom swap. Split the integral at each crossing, or integrate ∣f(x)−g(x)∣|f(x) - g(x)|.

When the region is easier to describe from left to right in y, write each curve as x in terms of y and use ∫cd(right−left)dy\int_c^d \left(\text{right} - \text{left}\right)dy.

If two rates are graphed, the area between them is the change in the difference of the amounts. When an inflow rate stays above an outflow rate, the area between them is how much the stored amount grows.

Trap A net integral is not an area

On [0,2][0, 2] the curves y=xy = x and y=x2y = x^2 cross at x=1x = 1. ∫02(x−x2) dx=−23\int_0^2 (x - x^2)\,dx = -\frac{2}{3}, a net value. The area is ∫01(x−x2) dx+∫12(x2−x) dx=16+56=1\int_0^1 (x - x^2)\,dx + \int_1^2 (x^2 - x)\,dx = \frac{1}{6} + \frac{5}{6} = 1.

Worked example Easy

What is the area of the region enclosed by y=2xy = 2x and y=x2y = x^2?

  1. 43\frac{4}{3}Answer
  2. 83\frac{8}{3}
  3. 44
  4. −43-\frac{4}{3}

How to solve it

  1. Set 2x=x22x = x^2: x(x−2)=0x(x - 2) = 0, so x=0x = 0 and x=2x = 2.
  2. At x=1x = 1, 2x=2>1=x22x = 2 > 1 = x^2, so the line is on top.
  3. ∫02(2x−x2) dx=[x2−x33]02=4−83=43\int_0^2 (2x - x^2)\,dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}.

Why each choice is right or wrong

  • A. Correct. The curves meet at x=0x = 0 and x=2x = 2, the line is on top, and ∫02(2x−x2) dx=4−83=43\int_0^2 (2x - x^2)\,dx = 4 - \frac{8}{3} = \frac{4}{3}.
  • B. Incorrect. 83\frac{8}{3} is ∫02x2 dx\int_0^2 x^2\,dx, the area under the parabola alone, not between the two curves.
  • C. Incorrect. 4 is ∫022x dx\int_0^2 2x\,dx, the area under the line alone; the area under the parabola must be subtracted.
  • D. Incorrect. −43-\frac{4}{3} subtracts top from bottom, x2−2xx^2 - 2x; an area is never negative.

Worked example Medium

What is the area of the region enclosed by y=6−x2y = 6 - x^2 and y=xy = x?

  1. 553\frac{55}{3}
  2. 223\frac{22}{3}
  3. 1256\frac{125}{6}Answer
  4. 956\frac{95}{6}

How to solve it

  1. Set 6−x2=x6 - x^2 = x: x2+x−6=0x^2 + x - 6 = 0, so (x+3)(x−2)=0(x + 3)(x - 2) = 0 and x=−3x = -3 or x=2x = 2.
  2. At x=0x = 0 the parabola is at 6 and the line at 0, so the parabola is on top.
  3. ∫−32(6−x−x2) dx=[6x−x22−x33]−32=223−(−272)=1256\int_{-3}^{2} (6 - x - x^2)\,dx = \left[6x - \frac{x^2}{2} - \frac{x^3}{3}\right]_{-3}^{2} = \frac{22}{3} - \left(-\frac{27}{2}\right) = \frac{125}{6}.

Why each choice is right or wrong

  • A. Incorrect. 553\frac{55}{3} is ∫−32(6−x2) dx\int_{-3}^{2} (6 - x^2)\,dx, which leaves out the line y = x.
  • B. Incorrect. 223\frac{22}{3} integrates only from 0 to 2, missing the part of the region to the left of the y-axis.
  • C. Correct. The curves meet at x=−3x = -3 and x=2x = 2, and ∫−32(6−x2−x) dx=1256\int_{-3}^{2} (6 - x^2 - x)\,dx = \frac{125}{6}.
  • D. Incorrect. 956\frac{95}{6} comes from ∫−32(6−x2+x) dx\int_{-3}^{2} (6 - x^2 + x)\,dx, which adds the line instead of subtracting it.

Explain Explain it

Task verbs: construct, calculate

Set up before you integrate. State the intersection points, say which curve is on top (test one point between them), then write the integral with its limits. A clear setup earns credit even if the arithmetic slips.

If a curve is a sideways parabola such as x=9−y2x = 9 - y^2, or the top or bottom curve changes partway across, try slicing in y. One integral in y can replace two in x.

Worked example Hard

What is the area of the region enclosed by the curve x=2y2−2x = 2y^2 - 2 and the line x=2y+2x = 2y + 2?

  1. 193\frac{19}{3}
  2. 99Answer
  3. 323\frac{32}{3}
  4. 33

How to solve it

  1. Set 2y2−2=2y+22y^2 - 2 = 2y + 2: y2−y−2=0y^2 - y - 2 = 0, so (y−2)(y+1)=0(y - 2)(y + 1) = 0 and y=−1y = -1 or y=2y = 2.
  2. At y=0y = 0 the line gives x=2x = 2 and the parabola x=−2x = -2, so the line is on the right.
  3. ∫−12(4+2y−2y2) dy=[4y+y2−2y33]−12=203−(−73)=9\int_{-1}^{2} (4 + 2y - 2y^2)\,dy = \left[4y + y^2 - \frac{2y^3}{3}\right]_{-1}^{2} = \frac{20}{3} - \left(-\frac{7}{3}\right) = 9.

Why each choice is right or wrong

  • A. Incorrect. 193\frac{19}{3} slices in x from 0 to 6 only, between y=x+22y = \sqrt{\frac{x + 2}{2}} and the line; it misses the part of the region for −2≤x≤0-2 \le x \le 0.
  • B. Correct. The curves meet at y=−1y = -1 and y=2y = 2, the line is on the right, and ∫−12((2y+2)−(2y2−2))dy=∫−12(4+2y−2y2) dy=9\int_{-1}^{2} \left((2y + 2) - (2y^2 - 2)\right)dy = \int_{-1}^{2} (4 + 2y - 2y^2)\,dy = 9.
  • C. Incorrect. 323\frac{32}{3} slices in x from −2-2 to 6 with the line as the bottom everywhere, but for −2<x<0-2 < x < 0 the bottom is the lower half of the parabola, not the line.
  • D. Incorrect. 3 is ∫−12(2y+2−2y2) dy\int_{-1}^{2} (2y + 2 - 2y^2)\,dy, which drops the −2-2 from the parabola, using x=2y2x = 2y^2 as the left boundary.

Construct Your turn

Write an integral for the area of the region enclosed by y=x2y = x^2 and y=2x+3y = 2x + 3. Then evaluate it. Show where the limits come from.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Finds the limits x=−1x = -1 and x=3x = 3 from where the curves meet
  • Writes the integrand as the line minus the parabola, ∫−13(2x+3−x2)dx\int_{-1}^{3} \left(2x + 3 - x^2\right)dx
  • Evaluates the area as 323\frac{32}{3}

Model answer

The curves meet where x2=2x+3x^2 = 2x + 3, so x2−2x−3=0x^2 - 2x - 3 = 0, (x−3)(x+1)=0(x - 3)(x + 1) = 0, and x=−1x = -1 or x=3x = 3. At x=0x = 0 the line is at 3 and the parabola at 0, so the line is on top. The area is ∫−13(2x+3−x2)dx=[x2+3x−x33]−13=9−(−53)=323\int_{-1}^{3} \left(2x + 3 - x^2\right)dx = \left[x^2 + 3x - \frac{x^3}{3}\right]_{-1}^{3} = 9 - \left(-\frac{5}{3}\right) = \frac{32}{3}.

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