# Mean Value Theorem for Integrals

In our discussions on derivatives, you learned about the Mean Value Theorem - an important theorem that claims that a function will take on its average value over an interval at least once. The Mean Value Theorem also has an application for integrals that is a consequence of the Mean Value Theorem and the Fundamental Theorem of Calculus.

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## Formula and meaning of the Mean Value Theorem for Integrals

The Mean Value Theorem for integrals states that if a function f is continuous on the closed interval [a, b], then there is a number c such that

${\int }_{a}^{b}f\left(x\right)dx=f\left(c\right)\left(b-a\right)$

Clearly, the left-hand side of the equation is the area under the curve of f on the interval (a, b). The right-hand side can be thought of as the area of a rectangle. So, the theorem states that the area under the curve is equal to the area of a rectangle with a width of the interval (b - a) and a height equal to the average value of the function f. Rearranging this equation to solve for f(c), the average value, we find

$f\left(c\right)=\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)dx$

Let us visualize the Mean Value Theorem for integrals geometrically.

The area under the curve of a function f on the interval [a, b] is equal to a rectangle with a width of b - a and a height of the average value of f, f(c) - StudySmarter Original

## Proof of the Mean Value Theorem for Integrals

Consider the definition of an antiderivative where

$F\left(x\right)={\int }_{a}^{x}f\left(t\right)dt$

By the Fundamental Theorem of Calculus

$F\text{'}\left(x\right)=f\left(x\right)$ and $F\left(b\right)-F\left(a\right)={\int }_{a}^{b}f\left(x\right)dx$

Because F is continuous over the closed interval [a, b] and differentiable over the open interval (a, b), we can apply the Mean Value Theorem, which says there is a number c such that $a and

$f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Using the outcomes of the Fundamental Theorem of Calculus

$f\left(c\right)=\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)dx$

## Examples of the Mean Value Theorem for Integrals

### Example 1

For the function $f\left(x\right)={x}^{2}+3x$ over the interval [1, 4], find the value c (the x-value where f(x) takes on its average value).

#### Step 1: Make sure f(x) is continuous over the closed interval

Since f(x) is a polynomial, we know it is continuous over the interval [1, 4].

#### Step 2: Evaluate the integral of f(x) over the given interval

$\begin{array}{rcl}{\int }_{1}^{4}{x}^{2}+3xdx& =& \left(\frac{{\left(4\right)}^{3}}{3}+\frac{3{\left(4\right)}^{2}}{2}\right)-\left(\frac{{\left(1\right)}^{3}}{3}+\frac{3{\left(1\right)}^{2}}{2}\right)\\ & =& 43.5\end{array}$

#### Step 3: Apply the Mean Value Theorem for integrals to find the average value of f(x) over the interval

$\begin{array}{rcl}f\left(c\right)& =& \frac{1}{4-1}{\int }_{1}^{4}{x}^{2}+3xdx\\ & =& \frac{1}{3}\left(43.5\right)\\ & =& 14.5\end{array}$

So, the average value that f(x) takes on is 14.5.

In Step 2, we found that the area under the curve is $43.5unit{s}^{2}$. To find the area of the rectangle, we multiply the width by the height.

$\left(4-1\right)\left(14.5\right)=43.5unit{s}^{2}$

Thus, the Mean Value Theorem for integrals holds.

#### Step 4: Find the x-value of f(c)

Since $f\left(c\right)=14.5$ and we want to find c, we can set f(x) equal to 14.5.

$14.5={x}^{2}+3x\phantom{\rule{0ex}{0ex}}0={x}^{2}+3x-14.5$

To solve for x, we apply the quadratic formula.

$\begin{array}{rcl}x& =& \frac{-b±\sqrt{{b}^{2}-4ac}}{2a}\\ x& =& \frac{-3±\sqrt{{3}^{2}-4\left(1\right)\left(-14.5\right)}}{2\left(1\right)}\\ x& =& \frac{-3+\sqrt{67}}{2}\approx 2.59and\\ x& =& \frac{-3-\sqrt{67}}{2}\approx -5.59\end{array}$

Since $-5.59$ is outside of the interval, $c\approx 2.59$.

### Example 2

For the function $f\left(x\right)=x+\mathrm{sin}\left(2x\right)$, find the x-value where f(x) takes on the average value over the interval $\left[0,2\mathrm{\pi }\right]$

#### Step 1: Make sure f(x) is continuous over the open interval

The function sin(x) is continuous everywhere.

#### Step 2: Evaluate the integral of f(x) over the given interval

$\begin{array}{rcl}{\int }_{0}^{2\mathrm{\pi }}x+\mathrm{sin}\left(2x\right)dx& =& \left(\frac{{\left(2\mathrm{\pi }\right)}^{2}}{2}-\frac{\mathrm{cos}\left(4\mathrm{\pi }\right)}{2}\right)-\left(\frac{{\left(0\right)}^{2}}{2}-\frac{\mathrm{cos}\left(0\right)}{2}\right)\\ & =& \frac{4{\mathrm{\pi }}^{2}}{2}-\frac{1}{2}-\left(0-\frac{1}{2}\right)\\ & =& 2{\mathrm{\pi }}^{2}-\frac{1}{2}+\frac{1}{2}\\ & =& 2{\mathrm{\pi }}^{2}\end{array}$

Use your knowledge of the unit circle to solve the trigonometric equations! Remember, $4\mathrm{\pi }$ is just a multiple of $2\mathrm{\pi }$.

#### Step 3: Apply the Mean Value Theorem for integrals to find the average value of f(x) over the interval

$\begin{array}{rcl}f\left(c\right)& =& \frac{1}{2\mathrm{\pi }-0}{\int }_{0}^{2\mathrm{\pi }}x+\mathrm{sin}\left(2x\right)dx\\ & =& \frac{1}{2\mathrm{\pi }}\left(2{\mathrm{\pi }}^{2}\right)\\ & =& \mathrm{\pi }\end{array}$

So, the average value that f(x) takes on is $\mathrm{\pi }$.

In Step 2, we found that the area under the curve is $2{\mathrm{\pi }}^{2}$ units2. To find the area of the rectangle, we multiply the width by the height.

$\left(2\mathrm{\pi }-0\right)\left(\mathrm{\pi }\right)=2{\mathrm{\pi }}^{2}$units2

Thus, the Mean Value Theorem for integrals holds.

#### Step 4: Find the x-value of f(c)

Since $f\left(c\right)=\mathrm{\pi }$ and we want to find c, we can set f(x) equal to $\mathrm{\pi }$.

$\mathrm{\pi }=\mathrm{x}+\mathrm{sin}\left(2\mathrm{x}\right)$

Solving this equation graphically, we find that $x=\mathrm{\pi }$.

## The Mean Value Theorem for Integrals Calculation

As a reminder

${\int }_{a}^{b}f\left(x\right)dx=f\left(c\right)\left(b-a\right)$

## Mean Value Theorem for Integrals - Key takeaways

• The Mean Value Theorem for integrals states that if a function f is continuous on the closed interval [a, b], then there is a number c such that

${\int }_{a}^{b}f\left(x\right)dx=f\left(c\right)\left(b-a\right)$

• Geometrically speaking, the area under the curve is equal to the area of a rectangle with a width of b - a and a height of the average value of f(x), f(c)

• The Mean Value Theorem for integrals is a consequence of the Mean Value Theorem for derivatives and the Fundamental Theorem of Calculus

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What is the mean value theorem for integrals?

The Mean Value Theorem for integrals states that if a function f is continuous on the closed interval [a, b], then the area under the curve is equal to the are of a rectangle with width b - a and height equal to the average value of the function f.

How do you use the Mean Value Theorem for integration?

To use the Mean Value Theorem, integrate the function over the given interval (a, b). Multiply the area under the curve by 1/(b-a) to find the average value over the given interval.

How do you find the values of C that satisfy the Mean Value Theorem for Integrals?

To find the value c, apply the Mean Value Theorem for integrals to find the function value at c. Then, set f(x) equal to f(c) and solve for x.

What is an example for the mean value theorem for integral?

A simple example of the Mean Value Theorem for integrals is the function f(x)=x over the interval [0, 1] has an average value of 1/2 at x = 1/2. This means that the area under the of f(x) over the interval [0, 1] is equal to the area of a rectangle with a width of 1 and a height of 1/2.

What is the formula and equation for finding the mean value theorem for integrals?

The formula for the Mean Value Theorem for integrals says that the definite integral from a to b is equal to f(c)(b - a) for some c value.

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