# Fundamental Theorem of Calculus: Parts 1 and 2 With Examples

> Learn both parts of the fundamental theorem of calculus: evaluate definite integrals with antiderivatives and differentiate integrals with variable limits.

Source: https://practicecalculus.com/blog/fundamental-theorem-of-calculus
Updated: 2026-09-28

## Key takeaways

- Part 2 evaluates definite integrals: if $F'= f$, then $\int_a^b f(x)\,dx = F(b) - F(a)$.
- Part 1 says integrating then differentiating gets you back where you started: $\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$.
- With a variable upper limit $u(x)$, use the chain rule: $\frac{d}{dx}\int_a^{u(x)} f(t)\,dt = f(u(x))\,u'(x)$.
- Definite integrals measure signed area, so area below the $x$-axis counts as negative.

Computing the area under $y = x^2$ with [Riemann sums](/blog/riemann-sums) takes a page of algebra and a limit. The **fundamental theorem of calculus** does it in one line: $\left[\frac{x^3}{3}\right]_0^2 = \frac{8}{3}$. It is the reason integrals are practical at all, and it connects the two halves of calculus into one subject.

## What the fundamental theorem of calculus says

In plain language: **differentiation and integration are inverse processes.** The theorem comes in two parts, one for each direction.

**Part 1 (derivative of an integral).** If $f$ is continuous on $[a, b]$, then the area function $g(x) = \int_a^x f(t)\,dt$ is differentiable and

$$\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$$

**Part 2 (evaluating integrals).** If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$ (so $F' = f$), then

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

We write $F(b) - F(a)$ as $\Big[F(x)\Big]_a^b$ for short.

Most textbooks number them this way, though a few swap the labels. What matters is recognizing which job you are doing: evaluating a number, or differentiating an integral.

## Why it works: area accumulates at a rate equal to the height

Picture the area under $y = f(t)$ from $a$ to $x$, and push $x$ a tiny bit to the right, by $h$. The extra area is a thin strip of width $h$ and height about $f(x)$, so

$$g(x + h) - g(x) \approx f(x)\cdot h \quad\Rightarrow\quad \frac{g(x+h) - g(x)}{h} \approx f(x)$$

As $h \to 0$ this becomes $g'(x) = f(x)$: **the rate at which area grows is the height of the curve.** That is part 1. Part 2 follows: $g$ is an antiderivative of $f$, any other antiderivative $F$ differs from it by a constant, and $\int_a^b f = g(b) - g(a) = F(b) - F(a)$.

## How to evaluate a definite integral (FTC part 2)

1. Find an antiderivative $F(x)$. Skip the $+C$, it cancels.
2. Compute $F(b)$, the top limit.
3. Compute $F(a)$, the bottom limit.
4. Subtract: $F(b) - F(a)$.

### Worked example 1: a polynomial

Evaluate $\int_0^2 x^2\,dx$.

**Step 1.** An antiderivative of $x^2$ is $\frac{x^3}{3}$.

**Step 2.** $F(2) = \frac{8}{3}$.

**Step 3.** $F(0) = 0$.

**Step 4.**

$$\int_0^2 x^2\,dx = \frac{8}{3} - 0 = \frac{8}{3}$$

This is exactly the limit of Riemann sums computed the long way, with no sigma notation needed.

### Worked example 2: roots and constants

Evaluate $\int_1^4 (3\sqrt{x} - 2)\,dx$.

**Step 1.** Write $3\sqrt{x} = 3x^{1/2}$. Its antiderivative is $3\cdot\frac{x^{3/2}}{3/2} = 2x^{3/2}$, and the antiderivative of $-2$ is $-2x$. So $F(x) = 2x^{3/2} - 2x$.

**Step 2.** $F(4) = 2(4)^{3/2} - 2(4) = 2(8) - 8 = 8$.

**Step 3.** $F(1) = 2(1) - 2(1) = 0$.

**Step 4.**

$$\int_1^4 (3\sqrt{x} - 2)\,dx = 8 - 0 = 8$$

### Worked example 3: the area under one arch of sine

Evaluate $\int_0^{\pi} \sin x\,dx$.

**Step 1.** An antiderivative of $\sin x$ is $-\cos x$.

**Steps 2 to 4.**

$$\int_0^{\pi}\sin x\,dx = \Big[-\cos x\Big]_0^{\pi} = (-\cos\pi) - (-\cos 0) = (1) - (-1) = 2$$

```graph
title: 'Area under one arch of y = sin(x) from 0 to π equals 2'
caption: 'The shaded region has area exactly $\int_0^{\pi}\sin x\,dx = 2$, found with one antiderivative and no rectangles.'
x: [-0.5, 3.8]
y: [-0.5, 1.5]
functions:
  - expr: sin(x)
    label: y = sin(x)
shade: {expr: sin(x), from: 0, to: 3.14159}
points:
  - [0, 0, '0']
  - [3.14159, 0, 'π']
```

The two minus signs are where most errors happen. Write the brackets out in full, as above, every time.

### Worked example 4: a logarithm

Evaluate $\int_1^e \frac{1}{x}\,dx$.

$$\int_1^e \frac{1}{x}\,dx = \Big[\ln|x|\Big]_1^e = \ln e - \ln 1 = 1 - 0 = 1$$

### Worked example 5: exponentials and a constant that does not vanish

Evaluate $\int_0^1 (e^x + 2x)\,dx$.

**Step 1.** An antiderivative is $F(x) = e^x + x^2$.

**Step 2.** $F(1) = e + 1$.

**Step 3.** $F(0) = e^0 + 0 = 1$. Watch this one: $e^0$ is $1$, not $0$, so the bottom limit does not disappear.

**Step 4.**

$$\int_0^1 (e^x + 2x)\,dx = (e + 1) - 1 = e \approx 2.718$$

## Signed area: when the integral is negative

A definite integral counts area **above** the $x$-axis as positive and area **below** as negative. For example,

$$\int_0^{2\pi}\sin x\,dx = \Big[-\cos x\Big]_0^{2\pi} = -1 - (-1) = 0$$

The arch above the axis (area $2$) cancels the arch below (area $2$). If a question asks for the **total area** rather than the integral, split at the zeros and add absolute values: $2 + 2 = 4$.

### Worked example 6: displacement versus distance traveled

The same idea appears in motion problems, where part 2 is often called the **net change theorem**: integrating a rate gives the total change. A particle moves with velocity $v(t) = 3t^2 - 6t$ m/s. Find its displacement and the total distance it travels from $t = 0$ to $t = 3$.

**Step 1. Displacement.** An antiderivative of $v$ is $t^3 - 3t^2$, so

$$\int_0^3 (3t^2 - 6t)\,dt = \Big[t^3 - 3t^2\Big]_0^3 = (27 - 27) - 0 = 0$$

The particle ends where it started.

**Step 2. Find where it turns around.** $v(t) = 3t(t - 2)$ is negative on $(0, 2)$ and positive on $(2, 3)$.

**Step 3. Distance.** Integrate $|v|$ by splitting at $t = 2$:

$$\int_0^2 -(3t^2 - 6t)\,dt = -\Big[t^3 - 3t^2\Big]_0^2 = -(8 - 12) = 4$$

$$\int_2^3 (3t^2 - 6t)\,dt = \Big[t^3 - 3t^2\Big]_2^3 = 0 - (-4) = 4$$

It moves $4$ m backward and then $4$ m forward, so the total distance is $8$ m even though the displacement is $0$.

## Derivatives of integrals (FTC part 1)

Part 1 lets you differentiate an integral without evaluating it, which matters when the integrand has no simple antiderivative.

### Worked example 7: x in the upper limit

Find $\frac{d}{dx}\int_1^x \sqrt{t^3 + 1}\,dt$.

There is no nice antiderivative of $\sqrt{t^3 + 1}$, and you do not need one. By part 1, replace $t$ with $x$:

$$\frac{d}{dx}\int_1^x \sqrt{t^3 + 1}\,dt = \sqrt{x^3 + 1}$$

The lower limit $1$ does not matter. Changing it only shifts the area function by a constant.

### Worked example 8: a function in the upper limit

Find $\frac{d}{dx}\int_0^{x^2}\cos t\,dt$.

Here the upper limit is $u = x^2$, so the chain rule adds a factor of $u' = 2x$:

$$\frac{d}{dx}\int_0^{x^2}\cos t\,dt = \cos(x^2)\cdot 2x$$

**Check by evaluating first:** $\int_0^{x^2}\cos t\,dt = \sin(x^2) - \sin 0 = \sin(x^2)$, and $\frac{d}{dx}\sin(x^2) = 2x\cos(x^2)$. It matches.

The general rule is

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

### Worked example 9: x in the lower limit

Find $\frac{d}{dx}\int_x^3 e^{t^2}\,dt$.

**Step 1. Flip the limits.** Swapping the limits changes the sign: $\int_x^3 e^{t^2}\,dt = -\int_3^x e^{t^2}\,dt$.

**Step 2. Apply part 1.**

$$\frac{d}{dx}\int_x^3 e^{t^2}\,dt = -e^{x^2}$$

If both limits depend on $x$, split the integral at a constant: $\frac{d}{dx}\int_{u(x)}^{v(x)} f(t)\,dt = f(v(x))\,v'(x) - f(u(x))\,u'(x)$.

## Quick reference

| Task                                    | Rule                                                     |
| --------------------------------------- | -------------------------------------------------------- |
| Evaluate $\int_a^b f(x)\,dx$            | $F(b) - F(a)$ where $F' = f$                             |
| $\frac{d}{dx}\int_a^x f(t)\,dt$         | $f(x)$                                                   |
| $\frac{d}{dx}\int_a^{u(x)} f(t)\,dt$    | $f(u(x))\,u'(x)$                                         |
| $\frac{d}{dx}\int_{x}^{a} f(t)\,dt$     | $-f(x)$                                                  |
| $\frac{d}{dx}\int_{u(x)}^{v(x)} f(t)\,dt$ | $f(v(x))\,v'(x) - f(u(x))\,u'(x)$                      |
| Swap limits                             | $\int_b^a f = -\int_a^b f$                               |

## Common mistakes with the fundamental theorem

1. **Subtracting in the wrong order.** It is $F(\text{top}) - F(\text{bottom})$. Reversing it flips the sign of your answer.
2. **Dropping parentheses with negatives.** In example 3, $(-\cos\pi) - (-\cos 0)$ needs every bracket. Writing $-\cos\pi - \cos 0$ gives $0$ instead of $2$.
3. **Forgetting the chain rule factor.** $\frac{d}{dx}\int_0^{x^2}\cos t\,dt$ is $2x\cos(x^2)$, not $\cos(x^2)$.
4. **Integrating when part 1 applies.** If the question asks for a derivative of an integral, you rarely need to find an antiderivative at all.
5. **Confusing integral with total area.** If the curve dips below the axis, $\int_a^b f$ is not the area. Split at the zeros.
6. **Applying it across a discontinuity.** $\int_{-1}^1 \frac{1}{x^2}\,dx$ is not $\left[-\frac{1}{x}\right]_{-1}^1 = -2$. The integrand blows up at $0$, so the theorem does not apply. That is an [improper integral](/blog/improper-integrals).

## Practice evaluating integrals

Once the theorem clicks, every definite integral becomes an antiderivative plus some careful arithmetic. Most errors are sign slips, so repetition with feedback is the fastest way to get reliable. The interactive lesson below checks each answer instantly and shows the full bracket work whenever a sign goes astray. For harder integrands, the next step is [u-substitution](/blog/u-substitution).

## Frequently asked questions

### What is the fundamental theorem of calculus in simple terms?

It says differentiation and integration undo each other. Part 1: the derivative of an accumulated area function is the original function. Part 2: to find the area under a curve, take any antiderivative and subtract its values at the two endpoints.

### What is the difference between FTC part 1 and part 2?

Part 1 is about derivatives of integrals: $\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$. Part 2 is about evaluating integrals: $\int_a^b f(x)\,dx = F(b) - F(a)$ for any antiderivative $F$. Part 1 is what guarantees antiderivatives exist; part 2 is what you use to compute.

### Do you need +C when evaluating a definite integral?

No. If you add $C$ to the antiderivative, it appears in both $F(b) + C$ and $F(a) + C$, and cancels when you subtract. Any antiderivative gives the same answer, so use the one with $C = 0$.

### How do you find the derivative of an integral with x in the upper limit?

Substitute the upper limit into the integrand and multiply by the derivative of the upper limit. For example $\frac{d}{dx}\int_0^{x^2}\cos t\,dt = \cos(x^2)\cdot 2x$. If the variable is in the lower limit, flip the limits first, which adds a minus sign.

### Can a definite integral be negative?

Yes. A definite integral adds area above the $x$-axis and subtracts area below it. For example $\int_{\pi}^{2\pi}\sin x\,dx = -2$ because the sine curve is below the axis there.