# Definite Integral Practice Problems with Answers

> 12 definite integral practice problems with answers and full solutions: polynomials, trig, symmetry, absolute value, geometry and integral properties.

Source: https://practicecalculus.com/practice-problems/definite-integrals
Updated: 2026-11-06

$$ \int_a^b f(x)\,dx = F(b) - F(a) $$

## At a glance

- **Problems:** 12, with answers and full solutions
- **Level:** Calculus 1 and AP Calculus AB, Unit 6
- **You need:** [Antiderivatives](/practice-problems/antiderivatives) and the [definite integral properties](/rules/definite-integral-properties)
- **Main idea:** Find an antiderivative, then subtract its value at the bottom limit from its value at the top
- **Practice with new numbers:** [Signed area lesson](/learn/calculus-1/integrals/signed-area)

## Key takeaways

- Evaluate as $F(b) - F(a)$, and put brackets around $F(a)$ so the minus sign reaches every term.
- A definite integral gives signed area: area below the $x$-axis counts as negative.
- Use symmetry, geometry and the integral properties to skip work when you can.

These **definite integral practice problems** start with direct evaluation and move on to symmetry, absolute values, geometry and the integral properties. Try each one first. The answers come first so you can check quickly, then the full solutions.

The tool for almost every problem is the [Fundamental Theorem of Calculus, Part 2](/theorems/fundamental-theorem-of-calculus-part-2):

$$\int_a^b f(x)\,dx = F(b) - F(a), \quad \text{where } F' = f$$

## Problems

**Direct evaluation**

1. $\displaystyle\int_0^3 (x^2 + 1)\,dx$
2. $\displaystyle\int_1^4 \sqrt{x}\,dx$
3. $\displaystyle\int_0^{\pi/2}\cos x\,dx$
4. $\displaystyle\int_1^{e}\frac{1}{x}\,dx$
5. $\displaystyle\int_{-1}^{2}(3x^2 - 2x)\,dx$
6. $\displaystyle\int_0^1 e^x\,dx$

**Shortcuts and splitting**

7. $\displaystyle\int_{-2}^{2}x^3\,dx$
8. $\displaystyle\int_0^3 |x - 1|\,dx$
9. Given $\int_0^5 f(x)\,dx = 8$ and $\int_0^2 f(x)\,dx = 3$, find $\int_2^5 f(x)\,dx$, $\int_5^2 2f(x)\,dx$ and $\int_2^5 \big(2f(x) + 1\big)\,dx$.
10. $\displaystyle\int_{-3}^{3}\sqrt{9 - x^2}\,dx$ (use geometry)

**Meaning**

11. Find the average value of $f(x) = x^2$ on $[0, 3]$.
12. Find $\int_0^{2\pi}\sin x\,dx$ and the total area between $y = \sin x$ and the $x$-axis on $[0, 2\pi]$.

## Answers

| Problem | Answer |
| --- | --- |
| 1 | $12$ |
| 2 | $\frac{14}{3}$ |
| 3 | $1$ |
| 4 | $1$ |
| 5 | $6$ |
| 6 | $e - 1$ |
| 7 | $0$ |
| 8 | $\frac{5}{2}$ |
| 9 | $5$, $-10$, $13$ |
| 10 | $\frac{9\pi}{2}$ |
| 11 | $3$ |
| 12 | Integral $0$, total area $4$ |

## Step-by-step solutions

**1.** An antiderivative is $\frac{x^3}{3} + x$:

$$\left[\frac{x^3}{3} + x\right]_0^3 = (9 + 3) - 0 = 12$$

**2.** Write $\sqrt{x} = x^{1/2}$, with antiderivative $\frac{2}{3}x^{3/2}$:

$$\frac{2}{3}\left(4^{3/2} - 1^{3/2}\right) = \frac{2}{3}(8 - 1) = \frac{14}{3}$$

**3.** An antiderivative of $\cos x$ is $\sin x$:

$$\sin\frac{\pi}{2} - \sin 0 = 1 - 0 = 1$$

**4.** An antiderivative of $\frac{1}{x}$ is $\ln x$ for $x > 0$:

$$\ln e - \ln 1 = 1 - 0 = 1$$

**5.** An antiderivative is $x^3 - x^2$. Keep the brackets around the bottom value:

$$\big(8 - 4\big) - \big((-1) - 1\big) = 4 - (-2) = 6$$

**6.** $e^x$ is its own antiderivative:

$$e^1 - e^0 = e - 1$$

**7.** $x^3$ is an odd function, and the interval $[-2, 2]$ is symmetric about $0$. The area on the left cancels the area on the right, so the integral is $0$. Check: $\left[\frac{x^4}{4}\right]_{-2}^{2} = 4 - 4 = 0$.

**8.** $|x - 1|$ changes formula at $x = 1$. Split there:

$$\int_0^1 (1 - x)\,dx + \int_1^3 (x - 1)\,dx = \frac{1}{2} + 2 = \frac{5}{2}$$

Both pieces are triangles: one with base $1$ and height $1$, one with base $2$ and height $2$.

**9.** The interval $[0, 5]$ splits into $[0, 2]$ and $[2, 5]$, so $\int_2^5 f(x)\,dx = 8 - 3 = 5$. Swapping limits flips the sign, so $\int_5^2 2f(x)\,dx = -2(5) = -10$. For the last one, $\int_2^5 1\,dx = 3$, so the answer is $2(5) + 3 = 13$. See the [definite integral properties](/rules/definite-integral-properties).

**10.** $y = \sqrt{9 - x^2}$ is the top half of a circle of radius $3$. Half of $\pi r^2$ is:

$$\frac{1}{2}\pi(3)^2 = \frac{9\pi}{2}$$

**11.** The average value is $\frac{1}{b-a}\int_a^b f(x)\,dx$:

$$\frac{1}{3}\int_0^3 x^2\,dx = \frac{1}{3}\cdot 9 = 3$$

**12.** The integral is $[-\cos x]_0^{2\pi} = -1 - (-1) = 0$. The hump above the axis cancels the hump below it. For total area, add the two humps as positive numbers. Each one has area $\int_0^{\pi}\sin x\,dx = 2$, so the total area is $4$.

## Common mistakes on these problems

- **Dropping brackets on $F(a)$.** In problem 5, writing $4 - 1 - 1$ instead of $4 - (-2)$ gives $2$ instead of $6$.
- **Treating signed area as total area.** Problem 12 shows they can be very different. Read the question for the word "area".
- **Integrating $|x - 1|$ as one piece.** Absolute values must be split where the inside changes sign.
- **Forgetting that swapped limits flip the sign.** $\int_5^2$ is the negative of $\int_2^5$.

## Keep practicing

For unlimited definite integrals with instant feedback, try the [signed area lesson](/learn/calculus-1/integrals/signed-area) and the [Fundamental Theorem, Part 2 lesson](/learn/calculus-1/integrals/fundamental-theorem-part-2). Then move on to [Fundamental Theorem problems](/practice-problems/fundamental-theorem-of-calculus) and [area between curves problems](/practice-problems/area-between-curves).

## Frequently asked questions

### Do you need +C in a definite integral?

No. The constant cancels when you subtract: $(F(b) + C) - (F(a) + C) = F(b) - F(a)$. A definite integral is a number.

### Can a definite integral be negative?

Yes. If more of the graph lies below the $x$-axis than above it on $[a, b]$, the integral is negative. If you want total area instead, integrate $|f(x)|$, as in problem 12.

### What happens if you swap the limits?

The sign flips: $\int_b^a f(x)\,dx = -\int_a^b f(x)\,dx$. Problem 9 uses this.