Mean Value Theorem for Integrals

The Mean Value Theorem for Integrals (MVTI) is an important result in calculus that connects:

• The average value of a function

• The definite integral of that function

It is sometimes called the Average Value Theorem.

Motivation

Imagine a continuous function f(x) on a closed interval [a,b].
The graph has areas above and below varying thickness.

The Mean Value Theorem for Integrals says:

There exists at least one point where the function reaches its average height over the interval.

This means the curve has a point where:

$$f(x)=f_{avg}$$​

Graphically, the horizontal line at the average height intersects the curve somewhere.

Average Value of a Function

The average value of a function $f(x)$ on [a,b] is defined as:

$$f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x)dx$$

This is similar to a regular average, but applied to continuous data.

Mean Value Theorem for Integrals (Statement)

If f(x) is continuous on a closed interval [a,b], then there exists at least one number c in the interval (a,b) such that:

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

This is the Mean Value Theorem for Integrals.

Geometric Interpretation

• $$\int_{a}^{b}f(x)dx$$ is the area under the curve.

• $$\frac{1}{b-a}$$​ spreads that area evenly across the interval—like flattening the curve into a rectangle of equal area.

The theorem says:

The function must reach the height of that rectangle at least once.

This matches intuition:
If a curve is continuous, it cannot “skip” its average height.

Connection to the Fundamental Theorem of Calculus

Let

$$F(x)=\int_{a}^{x}f(t)dt$$

By FTC Part 1:

$$f^\prime(x)=f(x)$$

Then applying the standard Mean Value Theorem (derivatives version):

$$F^\prime(x)=\frac{F(b)-F(a)}{b-a}$$​

Since $$F(a)=0$$ and $$F(b)=\int_{a}^{b}f(t)dt$$, we get:

$$f(c)=\frac{1}{b-a}\int_{a}^{b}f(t)dt$$

Thus, the Mean Value Theorem for derivatives gives us the Mean Value Theorem for integrals.

Example 1 (Simple Polynomial)

Find a value c such that MVTI holds for:

$$f(x)=x^{2}$$ on [0,3]

Step 1: Compute average value

$$f_{avg}=\frac{1}{3-0}\int_{0}^{3}x^{2}dx=\frac{1}{3}\cdot\frac{x^{3}}{3}\big|_{0}^{3}=\frac{1}{3}\cdot9=3$$

Step 2: Set $$f(c)=f_{avg}$$​

$$c^2=3 \longrightarrow c=\sqrt{3}$$​ 

Example 2 (Trigonometric)

Find c such that MVTI holds for:

$$f(x)=\sin(x) on [0,\pi]$$

1. Average value:

$$f_{avg}=\frac{1}{\pi}\int_{0}^{\pi}\sin(x)dx=\frac{1}{\pi}\cdot2=\frac{2}{\pi}$$​

2. Solve:

$$\sin{c}=\frac{2}{\pi} \longrightarrow c=\sin^{-1}\left(\frac{2}{\pi}\right)$$

Important Notes / Common Mistakes

✔ Must be continuous

If f(x) is not continuous, MVTI does not apply.

✔ Value c is guaranteed, but not unique

Some functions equal their average at multiple points.

✔ MVTI does not give a formula for c

It only guarantees existence.

Applications in Real Life

The MVTI helps to determine:

• Average temperature at some moment

• Average velocity corresponds to an instant speed

• Average electrical current corresponds to a real current value

• Average rate of growth equals actual rate at some time

• Economic average demand or revenue matches actual value at some price

Many physical systems use MVTI for mean measurements.