The objective of this question is to find the average value of the function over the given region which is a rectangle.
The average value of a bounded set of numbers is described as the total of the numbers divided by the number of numbers. In other words, a function’s average value is the average height of its graph. Among the most practical uses of the definite integral is that it describes the function’s average value, regardless of whether the function has an infinite number of values. The procedure of finding the average value of a function includes the usage of the FTC (Fundamental Theorem of Calculus), where the function is integrated over a bounded interval and is then divided by its length.
This computes the average height of a rectangle that will also encompass the exact area under the curve, which is the same as a function’s average value. Let $f(x)$ be a function over an interval $[a,b]$, then the average value of a function is defined as:
$f=\dfrac{1}{b-a}\int\limits_{a}^{b}f(x)dx$
Expert Answer
Let $A$ be the area of the region $R$, then the average value of the function over the region $R$ is given by:
$f=\dfrac{1}{A}\int\int_{R}f(x,y)dA$
Now, $A$ and $R$ can be defined as:
$A=2\times 5=10$ and $R=[-1,1]\times [0,5]$
With these values of $A$ and $R$, the above formula takes the form:
$f=\dfrac{1}{10}\int\limits_{-1}^{1}\int\limits_{0}^{5}x^2ydydx$
Next, keeping $x$ constant, integrate the above function with respect to $y$:
$f=\dfrac{1}{10}\int\limits_{-1}^{1}\left[\int\limits_{0}^{5}x^2ydy\right]dx$
$f=\dfrac{1}{10}\int\limits_{-1}^{1}\left[x^2\int\limits_{0}^{5}ydy\right]dx$
$f=\dfrac{1}{10}\int\limits_{-1}^{1}x^2\left[\dfrac{y^2}{2}\right]_{0}^{5}dx$
$f=\dfrac{1}{10}\int\limits_{-1}^{1}x^2\left[\dfrac{5^2}{2}-\dfrac{0^2}{2}\right]dx$
$f=\dfrac{1}{10}\int\limits_{-1}^{1}x^2\left[\dfrac{25}{2}\right]dx$
$f=\dfrac{1}{10}\times \dfrac{25}{2}\int\limits_{-1}^{1}x^2dx$
$f=\dfrac{5}{4}\left[\dfrac{x^3}{3}\right]_{-1}^{1}$
$f=\dfrac{5}{4}\left[\dfrac{(1)^3}{3}-\dfrac{(-1)^3}{3}\right]$
$f=\dfrac{5}{4}\left[\dfrac{1}{3}+\dfrac{1}{3}\right]$
$f=\dfrac{5}{4}\times \dfrac{2}{3}$
$f=\dfrac{5}{6}$
Example 1
Find the average value of the function $f(x)=(1+x)^2$ over the interval $-1\leq x \leq 0$.
Solution
The average value of a function over the interval $[a,b]$ is given by:
$f=\dfrac{1}{b-a}\int\limits_{a}^{b}f(x)dx$
where $a=-1, b=0$ and $f(x)=(1+x)^2$. Substitute these values in the above integral.
$f=\dfrac{1}{0-(-1)}\int\limits_{-1}^{0}(1+x)^2dx$
Next, expand $f(x)$ and then integrate:
$f=\dfrac{1}{0+1}\int\limits_{-1}^{0}(x^2+2x+1)dx$
$f=\int\limits_{-1}^{0}(x^2+2x+1)dx$
$f=\left[\dfrac{x^3}{3}+2\cdot \dfrac{x^2}{2}+x\right]_{-1}^{0}$
Apply the limits of integration as:
$f=\left[\dfrac{0}{3}+\dfrac{2(0)^2}{2}+0\right]-\left[-\dfrac{1}{3}+\dfrac{2}{2}-1\right]$
$f=0+\dfrac{1}{3}-1+1$
$f=\dfrac{1}{3}$
Example 2
Given the function $f(x)=\cos x$, find its average value on the interval $[0,\pi]$.
Solution
The average value of a function over the interval $[a,b]$ is given by:
$f=\dfrac{1}{b-a}\int\limits_{a}^{b}f(x)dx$
here, $a=-1, b=0$ and $f(x)=(1+x)^2$. Substitute these values in the above integral.
$f=\dfrac{1}{\pi-0}\int\limits_{0}^{\pi}\cos x dx$
$f=\dfrac{1}{\pi}[-\sin x]_{0}^{\pi}$
$f=-\dfrac{1}{\pi}[\sin \pi-\sin 0]$
$f=-\dfrac{1}{\pi}(0)$
$f=0$
Example 3
Given the function $f(x)=e^{2x}$, find its average value on the interval $[0,2]$.
Solution
Here, $a=0, b=2$
$f=\dfrac{1}{2-0}\int\limits_{0}^{2}e^{2x} dx$
$f=\dfrac{1}{2}\left[\dfrac{e^{2x}}{2}\right]_{0}^{2}$
$f=\dfrac{1}{2}\left[\dfrac{e^{4}}{2}-\dfrac{e^{0}}{2}\right]$
$f=\dfrac{1}{2}\left[\dfrac{e^{4}}{2}-\dfrac{1}{2}\right]$
$f=\dfrac{1}{4}(e^4-1)$