# 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.

Source: https://practicecalculus.com/practice-problems/area-between-curves
Updated: 2026-11-06

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

## At a glance

- **Problems:** 10, with answers and full solutions
- **Level:** Calculus 1, Calculus 2 and AP Calculus, Unit 8
- **You need:** [Definite integrals](/practice-problems/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](/learn/calculus-1/applications-of-integrals/area-between-curves)

## Key takeaways

- Set the curves equal to find where they meet. Those $x$-values are the limits.
- Test a point between the limits to see which curve is on top. If they cross inside the interval, split the integral there.
- When the region is bounded on the left and right, integrate in $y$: right curve minus left curve.

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](/blog/area-between-curves).

```graph
title: 'Region between y = x and y = x²'
caption: 'Problem 1: the line is on top between x = 0 and x = 1.'
x: [-0.5, 1.5]
y: [-0.5, 1.5]
functions:
  - expr: 'x'
    label: 'y = x'
  - expr: 'x^2'
    label: 'y = x²'
points:
  - [0, 0, '(0, 0)']
  - [1, 1, '(1, 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**

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

**Harder setups**

9. The region bounded by $x = y^2$ and $x = y + 2$
10. 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](/learn/calculus-1/applications-of-integrals/area-between-curves) and the [applications practice set](/learn/calculus-1/applications-of-integrals/applications-practice). Next, spin these regions around an axis in the [volumes of revolution problems](/practice-problems/volumes-of-revolution).

## 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 $x$.