## 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

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

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.

## 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 $a<c<b$ 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.

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

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.

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

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 $[0,2\mathrm{\pi}]$

#### 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 $2{\mathrm{\pi}}^{2}$ units^{2}. To find the area of the rectangle, we multiply the width by the height.

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

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

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

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

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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##### Frequently Asked Questions about Mean Value Theorem for Integrals

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 *f *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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