Definite Integral Practice Problems with Answers

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:

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

Step-by-step solutions

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

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

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

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

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

6. e^x is its own antiderivative:

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:

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.

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

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

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 and the Fundamental Theorem, Part 2 lesson. Then move on to Fundamental Theorem problems and area between curves problems.