practicecalculus
Applications of integrals

Area Between Curves Practice Problems with Answers

10 area between curves practice problems with answers and full solutions: finding intersections, top minus bottom, crossing curves and integrating in y.

Applications of integrals Updated 4 min read

At a glance

A = \int_a^b \big(\text{top}(x) - \text{bottom}(x)\big)\,dx
Problems
10, with answers and full solutions
Level
Calculus 1, Calculus 2 and AP Calculus, Unit 8
You need
Definite integrals and solving equations
Main idea
Integrate the top curve minus the bottom curve between the points where they meet
Practice with new numbers
Area between curves lesson
On this page
  1. Problems
  2. Answers
  3. Step-by-step solutions
  4. Common mistakes on these problems
  5. Keep practicing

These area between curves practice problems start with simple regions and move on to crossing curves and integrating with respect to y. Try each one first. The answers come first so you can check quickly, then the full solutions.

The formula:

A = \int_a^b \big(\text{top}(x) - \text{bottom}(x)\big)\,dx

Find where the curves meet, decide which one is on top, and integrate the difference. For a longer explanation with pictures, read the area between curves guide.

Region between y = x and y = x²
Problem 1: the line is on top between x = 0 and x = 1.

Problems

Two curves, one region

  1. y = x and y = x^2
  2. y = 4 - x^2 and y = 0
  3. y = 2x and y = x^2
  4. y = x^2 - 4 and y = 2 - x
  5. y = \sqrt{x} and y = x^2

Given limits

  1. y = \cos x and y = \sin x from x = 0 to x = \frac{\pi}{4}
  2. y = e^x and y = 1 from x = 0 to x = 1
  3. y = x and y = \frac{1}{x} from x = 1 to x = 2

Harder setups

  1. The region bounded by x = y^2 and x = y + 2
  2. The total area between y = x^3 and y = x on [-1, 1]

Answers

Problem Answer
1 \frac{1}{6}
2 \frac{32}{3}
3 \frac{4}{3}
4 \frac{125}{6}
5 \frac{1}{3}
6 \sqrt{2} - 1 \approx 0.414
7 e - 2 \approx 0.718
8 \frac{3}{2} - \ln 2 \approx 0.807
9 \frac{9}{2}
10 \frac{1}{2}

Step-by-step solutions

1. Set x = x^2: x = 0 or x = 1. At x = \frac{1}{2}, the line gives 0.5 and the parabola 0.25, so the line is on top:

\int_0^1(x - x^2)\,dx = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}

2. The parabola meets the x-axis at x = \pm 2:

\int_{-2}^{2}(4 - x^2)\,dx = \left[4x - \frac{x^3}{3}\right]_{-2}^{2} = \frac{16}{3} - \left(-\frac{16}{3}\right) = \frac{32}{3}

3. Set 2x = x^2: x = 0 or x = 2. At x = 1, the line is higher:

\int_0^2(2x - x^2)\,dx = 4 - \frac{8}{3} = \frac{4}{3}

4. Set x^2 - 4 = 2 - x: x^2 + x - 6 = 0, so x = -3 or x = 2. At x = 0, the line gives 2 and the parabola -4, so the line is on top:

\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}

5. The curves meet at x = 0 and x = 1, and \sqrt{x} is on top in between:

\int_0^1\left(x^{1/2} - x^2\right)dx = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}

6. On \left[0, \frac{\pi}{4}\right], \cos x \geq \sin x:

\int_0^{\pi/4}(\cos x - \sin x)\,dx = \Big[\sin x + \cos x\Big]_0^{\pi/4} = \sqrt{2} - 1

7. On [0, 1], e^x \geq 1:

\int_0^1(e^x - 1)\,dx = (e - 1) - (1 - 0) = e - 2

8. For x > 1, x > \frac{1}{x}:

\int_1^2\left(x - \frac{1}{x}\right)dx = \left(2 - \ln 2\right) - \left(\frac{1}{2} - 0\right) = \frac{3}{2} - \ln 2

9. Both curves are written as x in terms of y, so integrate in y. Set y^2 = y + 2: y = -1 or y = 2. At y = 0, the line gives x = 2 and the parabola x = 0, so the line is on the right:

\int_{-1}^{2}(y + 2 - y^2)\,dy = \left[\frac{y^2}{2} + 2y - \frac{y^3}{3}\right]_{-1}^{2} = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{9}{2}

10. The curves meet at x = -1, 0, 1. On [0, 1], x \geq x^3. On [-1, 0], x^3 \geq x. By symmetry the two pieces are equal:

2\int_0^1(x - x^3)\,dx = 2\left(\frac{1}{2} - \frac{1}{4}\right) = \frac{1}{2}

Integrating x - x^3 over [-1, 1] in one go gives 0, which is wrong.

Common mistakes on these problems

  • Bottom minus top. A negative answer is the giveaway. Test a point to find the top curve.
  • Not splitting at a crossing. In problem 10, one integral cancels the two regions and gives 0.
  • Using the wrong limits. The limits come from where the curves meet, not from the x-intercepts of one curve.
  • Mixing x and y. In problem 9, the integrand and the limits must both be in y.

Keep practicing

For unlimited regions with instant feedback, try the area between curves lesson and the applications practice set. Next, spin these regions around an axis in the volumes of revolution problems.

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Area between curves

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Frequently asked questions

How do you know which curve is on top?

Pick any x between the limits and plug it into both functions. The larger value is the top curve on that whole piece, as long as the curves do not cross inside it.

Can area between curves be negative?

No. Area is always positive. If you get a negative number, you subtracted in the wrong order. If curves cross, split the integral so that each piece is top minus bottom.

When should you integrate with respect to y?

When the curves are easier to write as x = f(y), or when integrating in x would need several pieces. Problem 9 is one integral in y but two in x.

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