# Fundamental Theorem of Calculus Practice Problems

> 12 Fundamental Theorem of Calculus practice problems with answers: Part 1 derivatives with variable limits, Part 2 evaluation and accumulation functions.

Source: https://practicecalculus.com/practice-problems/fundamental-theorem-of-calculus
Updated: 2026-11-06

$$ \frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x))\,g'(x) $$

## At a glance

- **Problems:** 12, with answers and full solutions
- **Level:** Calculus 1 and AP Calculus AB, Unit 6
- **You need:** The [chain rule](/rules/chain-rule) and basic antiderivatives
- **Main idea:** Part 1: plug the top limit into the integrand and multiply by its derivative. Part 2: $F(b) - F(a)$
- **Practice with new numbers:** [Fundamental Theorem, Part 1 lesson](/learn/calculus-1/integrals/fundamental-theorem-part-1)

## Key takeaways

- Part 1: the derivative of $\int_a^x f(t)\,dt$ is $f(x)$. With a function as the top limit, multiply by its derivative.
- If $x$ is in the bottom limit, flip the limits first, which adds a minus sign.
- An accumulation function increases where its integrand is positive and decreases where it is negative.

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:

$$\text{Part 1: } \frac{d}{dx}\int_a^{x} f(t)\,dt = f(x) \qquad \text{Part 2: } \int_a^b f(x)\,dx = F(b) - F(a)$$

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](/theorems/fundamental-theorem-of-calculus-part-1) and [Part 2](/theorems/fundamental-theorem-of-calculus-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

| Problem | Answer |
| --- | --- |
| 1 | $x^3 + 1$ |
| 2 | $2x\cos(x^2)$ |
| 3 | $-e^{x^2}$ |
| 4 | $\frac{1}{x}$ |
| 5 | $(1 + \sin^2 x)\cos x$ |
| 6 | $12$ |
| 7 | $2$ |
| 8 | $2$ |
| 9 | $\frac{\pi}{4}$ |
| 10 | Increasing for $\lvert x\rvert > 2$; local min at $x = 2$, $g(2) = -\frac{16}{3}$ |
| 11 | $F(2) = 0$, $F'(2) = 3$ |
| 12 | $2x\cos(x^2)$ |

## Step-by-step solutions

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

$$x^3 + 1$$

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

$$\cos(x^2)\cdot 2x = 2x\cos(x^2)$$

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

$$-e^{x^2}$$

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:

$$\frac{1}{x^2}\cdot 2x - \frac{1}{x}\cdot 1 = \frac{2}{x} - \frac{1}{x} = \frac{1}{x}$$

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 $\cos x$:

$$(1 + \sin^2 x)\cos x$$

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

$$(16 - 4) - (2 - 2) = 12$$

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

$$-\cos\pi - (-\cos 0) = 1 + 1 = 2$$

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

$$e^{\ln 3} - e^0 = 3 - 1 = 2$$

**9.** An antiderivative is $\arctan x$:

$$\arctan 1 - \arctan 0 = \frac{\pi}{4}$$

**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:

$$g(2) = \int_0^2 (t^2 - 4)\,dt = \frac{8}{3} - 8 = -\frac{16}{3}$$

**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:

$$g''(x) = \cos(x^2)\cdot 2x = 2x\cos(x^2)$$

## 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](/learn/calculus-1/integrals/fundamental-theorem-part-1), the [Part 2 lesson](/learn/calculus-1/integrals/fundamental-theorem-part-2) and the [accumulation functions lesson](/learn/calculus-1/integrals/accumulation-functions). For the big picture, read the [Fundamental Theorem guide](/blog/fundamental-theorem-of-calculus). Then move on to [u-substitution problems](/practice-problems/u-substitution).

## Frequently asked questions

### What is the difference between Part 1 and Part 2 of the Fundamental Theorem?

Part 1 says that differentiating an integral with a variable top limit gives back the integrand. Part 2 says you can evaluate a definite integral with any antiderivative, as $F(b) - F(a)$. Part 1 is about derivatives of integrals; Part 2 is about computing integrals.

### Why does the derivative of an integral not depend on the bottom limit?

Changing a constant bottom limit only adds or subtracts a fixed amount of area, which is a constant. Constants have derivative zero.

### What if both limits depend on x?

Split the integral at a constant, or use the rule directly: $\frac{d}{dx}\int_{h(x)}^{g(x)} f(t)\,dt = f(g(x))g'(x) - f(h(x))h'(x)$. Problem 4 shows this.