# Fundamental Theorem of Calculus Part 2: Evaluating Integrals

> FTC part 2: the definite integral of f from a to b equals F(b) − F(a) for any antiderivative F. Statement, graph, a key counterexample and worked examples.

Source: https://practicecalculus.com/theorems/fundamental-theorem-of-calculus-part-2
Updated: 2026-11-03

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

## At a glance

- **Also called:** Evaluation theorem, Newton-Leibniz formula
- **Hypotheses:** $f$ is continuous on $[a, b]$, and $F$ is any antiderivative of $f$ there ($F' = f$)
- **Conclusion:** $\int_a^b f(x)\,dx = F(b) - F(a)$
- **Notation:** $\big[F(x)\big]_a^b = F(b) - F(a)$
- **Net change form:** $\int_a^b F'(x)\,dx = F(b) - F(a)$
- **Course:** Calculus 1

## Key takeaways

- To evaluate $\int_a^b f(x)\,dx$, find any antiderivative $F$ and compute $F(b) - F(a)$.
- The $+ C$ cancels in the subtraction, so you can leave it out.
- $f$ must be continuous on the whole interval $[a, b]$. An infinite discontinuity inside makes the formula give nonsense.
- The integral of a rate of change is the net change: $\int_a^b F'(x)\,dx = F(b) - F(a)$.

The **fundamental theorem of calculus, part 2** (FTC 2), also called the evaluation theorem, says that a definite integral can be computed by finding an antiderivative and subtracting its values at the endpoints. It replaces the limit of Riemann sums with one line of algebra.

## Statement of FTC part 2

Suppose that

1. $f$ is **continuous** on $[a, b]$, and
2. $F$ is **any antiderivative** of $f$ on $[a, b]$, so $F'(x) = f(x)$.

Then

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

The right side is often written $\big[F(x)\big]_a^b$ or $F(x)\Big|_a^b$.

## What FTC 2 means

The definite integral adds up $f(x)\,dx$ across the interval. If $f$ is the rate of change of $F$, adding up all the small changes $F'(x)\,dx$ gives the total change in $F$, which is $F(b) - F(a)$. This is the **net change theorem**: integrate velocity to get displacement, integrate a flow rate to get the total volume.

```graph
title: 'FTC part 2: area under y = sin(x) from 0 to π equals 2'
caption: 'With antiderivative $F(x) = -\cos x$, the shaded area is $F(\pi) - F(0) = 1 - (-1) = 2$.'
x: [-0.5, 3.6]
y: [-0.5, 1.4]
functions:
  - expr: sin(x)
    label: y = sin(x)
shade: {expr: sin(x), from: 0, to: 3.14159}
vertical:
  - {x: 3.14159, label: 'x = π'}
```

## Why continuity matters

Consider $\int_{-1}^{1}\frac{1}{x^2}\,dx$. An antiderivative of $x^{-2}$ is $-\frac{1}{x}$, and blindly applying the formula gives

$$\left[-\frac{1}{x}\right]_{-1}^{1} = -1 - 1 = -2$$

But $\frac{1}{x^2} > 0$ everywhere, so its integral cannot be negative. The mistake: $\frac{1}{x^2}$ is not continuous on $[-1, 1]$ because it blows up at $x = 0$, so FTC 2 does not apply. Treated correctly as an [improper integral](/blog/improper-integrals), it diverges to $\infty$. Always check for discontinuities inside the interval before plugging in.

## Worked examples

**Example 1: a polynomial.** Evaluate $\int_1^3 (2x + 1)\,dx$.

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

**Step 2.** $F(3) = 9 + 3 = 12$ and $F(1) = 1 + 1 = 2$.

**Step 3.** $\int_1^3 (2x + 1)\,dx = 12 - 2 = 10$.

Check with geometry: the region is a trapezoid with parallel sides $f(1) = 3$ and $f(3) = 7$ and width $2$, so its area is $\frac{3 + 7}{2}\cdot 2 = 10$.

**Example 2: a trig integral.** Evaluate $\int_0^\pi \sin x\,dx$.

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

This is the area under one arch of the sine curve, shown in the graph above.

**Example 3: a logarithm.** Evaluate $\int_1^e \frac{1}{x}\,dx$.

$\frac{1}{x}$ is continuous on $[1, e]$, and an antiderivative is $\ln x$. So

$$\int_1^e \frac{1}{x}\,dx = \ln e - \ln 1 = 1 - 0 = 1$$

**Example 4: net change.** A tank fills at $r(t) = 3t^2$ liters per minute. Between $t = 1$ and $t = 2$ minutes the volume increases by

$$\int_1^2 3t^2\,dt = \big[t^3\big]_1^2 = 8 - 1 = 7 \text{ liters}$$

## Common mistakes

1. **Subtracting in the wrong order.** It is $F(b) - F(a)$, upper limit first.
2. **Dropping parentheses.** With $F(x) = x^2 - 3x$, write $F(b) - F(a) = (b^2 - 3b) - (a^2 - 3a)$. Losing the parentheses flips signs.
3. **Ignoring discontinuities.** As shown above, an integrand that blows up inside $[a, b]$ needs improper integral techniques.
4. **Confusing area with signed area.** FTC 2 gives the signed area: regions below the $x$-axis count as negative. For total area, split where $f$ changes sign, as in [area between curves](/blog/area-between-curves).

## How FTC 2 connects to FTC 1

[FTC part 1](/theorems/fundamental-theorem-of-calculus-part-1) says $G(x) = \int_a^x f(t)\,dt$ is an antiderivative of $f$. Any other antiderivative differs from $G$ by a constant (a consequence of the [mean value theorem](/theorems/mean-value-theorem)), so $F(b) - F(a) = G(b) - G(a) = \int_a^b f(t)\,dt$. That is the whole proof of part 2.

To find antiderivatives, see [antiderivatives](/blog/antiderivatives) and [u-substitution](/blog/u-substitution). For both parts of the theorem with more examples, read the [fundamental theorem of calculus guide](/blog/fundamental-theorem-of-calculus).

## Frequently asked questions

### What does the second fundamental theorem of calculus say?

If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$, then $\int_a^b f(x)\,dx = F(b) - F(a)$. A definite integral can be computed from an antiderivative instead of a limit of Riemann sums.

### Why don't you need the +C in a definite integral?

If you use $F(x) + C$, the result is $(F(b) + C) - (F(a) + C) = F(b) - F(a)$. The constant cancels, so any antiderivative gives the same answer.

### Can the fundamental theorem of calculus give a wrong answer?

Only if you ignore its hypothesis. For $\int_{-1}^{1}\frac{1}{x^2}\,dx$ the naive calculation gives $-2$, which is impossible for a positive integrand. The integrand is not continuous at $0$, so the theorem does not apply and the integral actually diverges.

### Is the fundamental theorem of calculus part 1 or part 2 used to evaluate integrals?

Part 2. Part 1 is about differentiating an integral with a variable limit, while part 2 is the $F(b) - F(a)$ formula used to evaluate definite integrals.