Fundamental Theorem of Calculus Practice Problems

These Fundamental Theorem of Calculus practice problems cover both parts of the theorem: taking derivatives of integrals with variable limits, evaluating definite integrals, and reading accumulation functions. Try each one first. The answers come first, then the full solutions.

The two parts:

With the chain rule, Part 1 becomes \frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x))\,g'(x). For the reasoning, see Part 1 and Part 2.

Problems

Part 1: derivatives of integrals

1. \displaystyle\frac{d}{dx}\int_1^{x}(t^3 + 1)\,dt
2. \displaystyle\frac{d}{dx}\int_0^{x^2}\cos t\,dt
3. \displaystyle\frac{d}{dx}\int_x^{5}e^{t^2}\,dt
4. \displaystyle\frac{d}{dx}\int_x^{x^2}\frac{1}{t}\,dt for x > 0
5. \displaystyle\frac{d}{dx}\int_0^{\sin x}(1 + t^2)\,dt

Part 2: evaluating

6. \displaystyle\int_1^{2}(6x^2 - 2)\,dx
7. \displaystyle\int_0^{\pi}\sin x\,dx
8. \displaystyle\int_0^{\ln 3}e^x\,dx
9. \displaystyle\int_0^{1}\frac{1}{1 + x^2}\,dx

Accumulation functions

10. Let g(x) = \int_0^{x}(t^2 - 4)\,dt. Where is g increasing? Find the local minimum of g and its value.
11. Let F(x) = \int_2^{x}\sqrt{t^2 + 5}\,dt. Find F(2) and F'(2).
12. Let g(x) = \int_0^{x}\sin(t^2)\,dt. Find g''(x).

Answers

Step-by-step solutions

1. The top limit is plain x, so replace t with x. The bottom limit 1 does not matter:

2. The top limit is x^2, so plug it in and multiply by its derivative 2x:

3. The variable is in the bottom limit. Flip the limits to get -\int_5^{x}e^{t^2}\,dt, then use Part 1:

Notice that e^{t^2} has no elementary antiderivative, so Part 1 is the only way to do this.

4. Top minus bottom, each with its chain rule factor:

Check: the integral equals \ln(x^2) - \ln x = \ln x, whose derivative is \frac{1}{x}.

5. Plug \sin x into 1 + t^2 and multiply by \cos x:

6. An antiderivative is 2x^3 - 2x:

7. An antiderivative of \sin x is -\cos x:

8. Since e^{\ln 3} = 3:

9. An antiderivative is \arctan x:

10. By Part 1, g'(x) = x^2 - 4. This is positive when |x| > 2, so g increases on (-\infty, -2) and (2, \infty). At x = 2, g' changes from negative to positive, so g has a local minimum there:

11. When the limits match, the integral is 0, so F(2) = 0. By Part 1, F'(x) = \sqrt{x^2 + 5}, so F'(2) = \sqrt{9} = 3.

12. First g'(x) = \sin(x^2). Differentiate again with the chain rule:

Common mistakes on these problems

• Forgetting the chain rule factor. In problem 2, \cos(x^2) alone is wrong. Multiply by the derivative of the top limit.
• Ignoring a variable bottom limit. In problem 3 the minus sign comes from flipping the limits.
• Integrating first when you cannot. e^{t^2} and \sin(t^2) have no elementary antiderivatives. Part 1 lets you skip that step.
• Confusing g and g'. In problem 10, g increases where the integrand is positive, not where g itself is positive.

Keep practicing

For unlimited problems with instant feedback, try the Fundamental Theorem, Part 1 lesson, the Part 2 lesson and the accumulation functions lesson. For the big picture, read the Fundamental Theorem guide. Then move on to u-substitution problems.