Mean Value Theorem for Integrals

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.

$$\begin{gathered} 14.5=x^2+3x \\ 0=x^2+3x-14.5 \end{gathered}$$

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, $c\approx 2.59$.

For the function $f\left(x\right)=x+\sin \left(2x\right)$, 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

$\begin{array}{rcl}{\int }_0^{2\mathrm{\pi }}x+\sin \left(2x\right)dx& =& \left(\frac{{\left(2\mathrm{\pi }\right)}^2}{2}-\frac{\cos \left(4\mathrm{\pi }\right)}{2}\right)-\left(\frac{{\left(0\right)}^2}{2}-\frac{\cos \left(0\right)}{2}\right)\\ & =& \frac{4{\mathrm{\pi }}^2}{2}-\frac{1}{2}-\left(0-\frac{1}{2}\right)\\ & =& 2{\mathrm{\pi }}^2-\frac{1}{2}+\frac{1}{2}\\ & =& 2{\mathrm{\pi }}^2\end{array}$

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

$\begin{array}{rcl}f\left(c\right)& =& \frac{1}{2\mathrm{\pi }-0}{\int }_0^{2\mathrm{\pi }}x+\sin \left(2x\right)dx\\ & =& \frac{1}{2\mathrm{\pi }}\left(2{\mathrm{\pi }}^2\right)\\ & =& \mathrm{\pi }\end{array}$

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.

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

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

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