The Mean Value Theorem for integrals

Levelling the area

Imagine the region under a curve is made of water held between two glass walls. Remove the curve and let the water settle. It forms a flat rectangle with the same area. How high does the water sit?

Drag the ochre level up and down. When the rectangle's area matches the area under the curve, you have found the average value of the function, and the theorem says the curve reaches that exact height somewhere in the interval.

Drag the ochre level. The rectangle has width b − a and your chosen height.

Function
Area under the curve
Rectangle area
Average value
Where f(c) equals it

The average value

To average a list of numbers, add them and divide by how many there are. A function on an interval has infinitely many values, so the sum becomes an integral and the count becomes the length of the interval:

f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)\,dx.

Multiply both sides by b-a and the formula says exactly what the water picture shows: a rectangle of height f_{\text{avg}} and width b-a has the same area as the region under the curve.

The theorem

\text{If } f \text{ is continuous on } [a,b], \text{ then } f(c)=\frac{1}{b-a}\int_a^b f(x)\,dx \text{ for some } c \text{ in } [a,b].

Why should this hold? The level of the water can be no higher than the curve's highest point and no lower than its lowest point. A continuous function takes every value between its lowest and highest, so by the Intermediate Value Theorem it must pass through the level somewhere.

The name is no accident. Write F(x)=\int_a^x f(t)\,dt. Then F'=f, and the Mean Value Theorem for derivatives applied to F gives

F'(c)=\frac{F(b)-F(a)}{b-a}\quad\Longleftrightarrow\quad f(c)=\frac{1}{b-a}\int_a^b f(x)\,dx.

It is the same theorem, read once as slopes and once as areas.

An example

For f(x)=x^2 on [0,3], the integral is 9, so f_{\text{avg}}=\tfrac93=3. Solving c^2=3 in the interval gives c=\sqrt3\approx1.732.

Solve it step by step

Find the integral, the average value, and the point where the function reaches it. The score counts steps you answer correctly on the first try.

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    Common mistakes

    Forgetting to divide

    The integral is the area. The average value is the area divided by the width b-a.

    Averaging the endpoints

    \tfrac{f(a)+f(b)}{2} is the average of two values, not of the whole function. For x^2 on [0,3] it gives 4.5 instead of 3.

    Keeping a c outside the interval

    Solving f(c)=f_{\text{avg}} can give several solutions. Only those in [a,b] count.