Question

How do you find the average value of the function for $f\left( x \right)={{e}^{x}}-2x,0\le x\le 2?$

Answer 1

Hint: The average value is the mean value. The average value of a function $f$ defined in the interval$\left[ a,b \right]$ is given by the formula $\overline{f\left( x \right)}=\dfrac{1}{b-a}\int\limits_{a}^{b}{f\left( x \right)}dx.$ We will compare the given values with the formula.

Complete step by step answer:
Let us consider the given function defined as $f\left( x \right)={{e}^{x}}-2x.$
This function is defined for all values of $x$ in the interval $\left[ 0,2 \right].$
We are asked to find the average value of the given function.
We know that the average value of a function $f\left( x \right)$ over the interval $\left[ a,b \right]$ is given by the formula $\overline{f\left( x \right)}=\dfrac{1}{b-a}\int\limits_{a}^{b}{f\left( x \right)}dx.$
We are going to compare the values to find the average value of the given function over the given interval.
We will get $a=0$ and $b=2,$ the interval is $\left[ 0,2 \right].$
Also, we know that $f\left( x \right)={{e}^{x}}-2x.$
Therefore, the average value of the given function is obtained by substituting these values in the formula for finding the average value of a function.
So, after applying the values, we will get $\overline{f\left( x \right)}=\dfrac{1}{b-a}\int\limits_{a}^{b}{f\left( x \right)}dx=\dfrac{1}{2-0}\int\limits_{0}^{2}{\left( {{e}^{x}}-2x \right)}dx.$
We will get the following $\overline{f\left( x \right)}=\dfrac{1}{b-a}\int\limits_{a}^{b}{f\left( x \right)}dx=\dfrac{1}{2}\int\limits_{0}^{2}{\left( {{e}^{x}}-2x \right)}dx=\dfrac{1}{2}\int\limits_{0}^{2}{{{e}^{x}}}dx+\dfrac{1}{2}\int\limits_{0}^{2}{2x}dx.$
We know that $\int{{{e}^{x}}={{e}^{x}}.}$
Therefore, the first integral will become $\int\limits_{0}^{2}{{{e}^{x}}}dx=\left[ {{e}^{x}} \right]_{0}^{2}={{e}^{2}}-{{e}^{0}}={{e}^{2}}-1.$
When we use the linearity property, the second integral will become $\int\limits_{0}^{2}{2x}dx=2\int\limits_{0}^{2}{x}dx=2\left[ \dfrac{{{x}^{2}}}{2} \right]_{0}^{2}=2\left( \dfrac{{{2}^{2}}}{2}-\dfrac{0}{2} \right)=2\left( \dfrac{4}{2}-0 \right)=2\times \dfrac{4}{2}=4.$
Now, we will apply these two values in the obtained equation instead of the terms with the symbol of integration.
We will get $\overline{f\left( x \right)}=\dfrac{1}{2}\int\limits_{0}^{2}{{{e}^{x}}}dx+\dfrac{1}{2}\int\limits_{0}^{2}{2x}dx=\dfrac{1}{2}\left( {{e}^{2}}-1 \right)+\dfrac{1}{2}\times 4.$
We will get $\overline{f\left( x \right)}=\dfrac{1}{2}{{e}^{2}}-1\times \dfrac{1}{2}+\dfrac{1}{2}\times 4=\dfrac{1}{2}{{e}^{2}}-\dfrac{1}{2}+2=\dfrac{1}{2}{{e}^{2}}+\dfrac{3}{2}.$
We know that ${{e}^{2}}=7.3891.$
Therefore, we will get $\dfrac{1}{2}{{e}^{2}}-\dfrac{3}{2}=\dfrac{1}{2}\left( {{e}^{2}}-3 \right)=\dfrac{1}{2}\left( 7.3891-3 \right)=\dfrac{1}{2}\times 4.3891.$
Hence the average value of the function is $2.1945.$

Note: By linearity property of the integration, we will get $\int\limits_{a}^{b}{\left( px+qy \right)}dx=p\int\limits_{a}^{b}{x}dx+q\int\limits_{a}^{b}{y}dx.$ Also, we have $k\int\limits_{a}^{b}{\left( x+y \right)}dx=k\int\limits_{a}^{b}{x}dx+k\int\limits_{a}^{b}{y}dx.$ We know that $e=2.71828.$