# Mean Value Theorem for Integrals: Statement and Examples

> The mean value theorem for integrals: a continuous function on [a, b] equals its average value at some point c. Statement, graph, counterexample, examples.

Source: https://practicecalculus.com/theorems/mean-value-theorem-for-integrals
Updated: 2026-11-03

$$ f(c) = \frac{1}{b - a} \int_a^b f(x)\,dx $$

## At a glance

- **Hypotheses:** $f$ is continuous on $[a, b]$
- **Conclusion:** Some $c$ in $[a, b]$ has $f(c) = \frac{1}{b - a}\int_a^b f(x)\,dx$
- **Average value:** $f_{\text{avg}} = \frac{1}{b - a}\int_a^b f(x)\,dx$
- **Geometric meaning:** A rectangle of height $f(c)$ over $[a, b]$ has the same area as the region under $f$
- **Related to:** The mean value theorem for derivatives, applied to $F(x) = \int_a^x f(t)\,dt$
- **Course:** Calculus 1

## Key takeaways

- A continuous function actually reaches its average value somewhere on the interval.
- The average value of $f$ on $[a, b]$ is $\frac{1}{b - a}\int_a^b f(x)\,dx$.
- To find $c$, compute the average value, then solve $f(c) = f_{\text{avg}}$ and keep solutions in $[a, b]$.
- Continuity is essential: a function with a jump can skip right over its average value.

The **mean value theorem for integrals** says a continuous function hits its own average value at least once. If you flatten the area under the curve into a rectangle of the same width, the rectangle's height is a value the function actually takes.

## Statement of the mean value theorem for integrals

If $f$ is **continuous** on the closed interval $[a, b]$, then there exists a number $c$ in $[a, b]$ such that

$$f(c) = \frac{1}{b - a}\int_a^b f(x)\,dx$$

or equivalently

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

The quantity on the right of the first equation is the **average value** of $f$ on $[a, b]$, written $f_{\text{avg}}$.

## What the theorem means

The area under $y = f(x)$ from $a$ to $b$ equals the area of a rectangle with the same base, $b - a$, and height $f_{\text{avg}}$. The theorem says that height is not just some number: the curve passes through it. Picture water in a tank with a wavy surface. When the water settles flat, its level is the average height, and the original wavy surface was at that level somewhere.

```graph
title: 'Mean value theorem for integrals: y = x² on [0, 3] reaches its average value 3 at x = √3'
caption: 'The area under $y = x^2$ on $[0, 3]$ is $9$, so the average value is $\frac{9}{3} = 3$. The curve equals $3$ at $c = \sqrt{3} \approx 1.732$.'
x: [-0.3, 3.3]
y: [-0.5, 9.5]
functions:
  - expr: x^2
    label: y = x²
  - expr: '3'
    label: average value y = 3
    dashed: true
    domain: [0, 3]
shade: {expr: x^2, from: 0, to: 3}
points:
  - [1.732, 3, 'c = √3']
```

## Why continuity matters

Let $f(x) = 0$ for $0 \le x < 1$ and $f(x) = 1$ for $1 \le x \le 2$. Its integral over $[0, 2]$ is $1$, so its average value is $\frac{1}{2}$. But $f$ only takes the values $0$ and $1$, never $\frac{1}{2}$. The jump at $x = 1$ lets the function skip over its average. With continuity, the [intermediate value theorem](/theorems/intermediate-value-theorem) rules this out.

## Why the theorem is true

By the [extreme value theorem](/theorems/extreme-value-theorem), $f$ has a minimum $m$ and maximum $M$ on $[a, b]$. Since $m \le f(x) \le M$,

$$m(b - a) \le \int_a^b f(x)\,dx \le M(b - a) \implies m \le f_{\text{avg}} \le M$$

So $f_{\text{avg}}$ lies between two values that $f$ attains, and the intermediate value theorem gives a $c$ with $f(c) = f_{\text{avg}}$.

## Worked examples

**Example 1: $f(x) = x^2$ on $[0, 3]$.**

**Step 1.** Average value: $\frac{1}{3 - 0}\int_0^3 x^2\,dx = \frac{1}{3}\left[\frac{x^3}{3}\right]_0^3 = \frac{1}{3}\cdot 9 = 3$.

**Step 2.** Solve $c^2 = 3$, so $c = \pm\sqrt{3}$.

**Step 3.** Only $c = \sqrt{3} \approx 1.732$ lies in $[0, 3]$.

**Example 2: $f(x) = \sin x$ on $[0, \pi]$.**

**Step 1.** $f_{\text{avg}} = \frac{1}{\pi}\int_0^\pi \sin x\,dx = \frac{1}{\pi}\cdot 2 = \frac{2}{\pi} \approx 0.637$.

**Step 2.** Solve $\sin c = \frac{2}{\pi}$. On $[0, \pi]$ there are two solutions:

$$c = \arcsin\frac{2}{\pi} \approx 0.690 \qquad \text{and} \qquad c = \pi - \arcsin\frac{2}{\pi} \approx 2.452$$

The theorem promises at least one; here there are two.

**Example 3: a linear function.** For $f(x) = 2x + 1$ on $[0, 4]$, $\int_0^4 (2x + 1)\,dx = \big[x^2 + x\big]_0^4 = 20$, so $f_{\text{avg}} = \frac{20}{4} = 5$. Solving $2c + 1 = 5$ gives $c = 2$, the midpoint. For any linear function the average value is reached at the midpoint.

## Common mistakes

1. **Forgetting to divide by $b - a$.** The integral is the area; the average value is the area divided by the width.
2. **Keeping solutions outside the interval.** In Example 1, $c = -\sqrt{3}$ must be discarded.
3. **Confusing it with the MVT for derivatives.** That theorem matches $f'(c)$ to the average slope of $f$. This one matches $f(c)$ to the average value of $f$.

## Related theorems

This theorem is the [mean value theorem](/theorems/mean-value-theorem) applied to $F(x) = \int_a^x f(t)\,dt$, using [FTC part 1](/theorems/fundamental-theorem-of-calculus-part-1) for $F' = f$ and [FTC part 2](/theorems/fundamental-theorem-of-calculus-part-2) to evaluate the integral. For more on integrals as accumulated area, see [Riemann sums](/blog/riemann-sums) and the [fundamental theorem of calculus guide](/blog/fundamental-theorem-of-calculus).

## Frequently asked questions

### What is the mean value theorem for integrals?

If $f$ is continuous on $[a, b]$, there is a number $c$ in $[a, b]$ with $f(c) = \frac{1}{b - a}\int_a^b f(x)\,dx$. In other words, $f$ takes on its average value at least once.

### How do you find the average value of a function?

Integrate the function over the interval and divide by the length of the interval: $f_{\text{avg}} = \frac{1}{b - a}\int_a^b f(x)\,dx$. For $x^2$ on $[0, 3]$ this is $\frac{1}{3}\cdot 9 = 3$.

### How is the mean value theorem for integrals related to the MVT for derivatives?

Let $F(x) = \int_a^x f(t)\,dt$. Applying the mean value theorem to $F$ gives $F'(c) = \frac{F(b) - F(a)}{b - a}$, and since $F' = f$ this is exactly $f(c) = \frac{1}{b - a}\int_a^b f(x)\,dx$.

### Is c in the open or closed interval?

Textbooks usually state $c \in [a, b]$, which is what the proof with the extreme and intermediate value theorems gives directly. A slightly sharper argument shows $c$ can always be chosen in the open interval $(a, b)$.