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

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

Explanation

Solution

The average value is the mean value. The average value of a function ff defined in the interval[a,b]\left[ a,b \right] is given by the formula f(x)‾=1b−a∫abf(x)dx.\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(x)=ex−2x.f\left( x \right)={{e}^{x}}-2x.
This function is defined for all values of xx in the interval [0,2].\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(x)f\left( x \right) over the interval [a,b]\left[ a,b \right] is given by the formula f(x)‾=1b−a∫abf(x)dx.\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=0a=0 and b=2,b=2, the interval is [0,2].\left[ 0,2 \right].
Also, we know that f(x)=ex−2x.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 f(x)‾=1b−a∫abf(x)dx=12−0∫02(ex−2x)dx.\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 f(x)‾=1b−a∫abf(x)dx=12∫02(ex−2x)dx=12∫02exdx+12∫022xdx.\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 ∫ex=ex.\int{{{e}^{x}}={{e}^{x}}.}
Therefore, the first integral will become ∫02exdx=[ex]02=e2−e0=e2−1.\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 ∫022xdx=2∫02xdx=2[x22]02=2(222−02)=2(42−0)=2×42=4.\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 f(x)‾=12∫02exdx+12∫022xdx=12(e2−1)+12×4.\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 f(x)‾=12e2−1×12+12×4=12e2−12+2=12e2+32.\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 e2=7.3891.{{e}^{2}}=7.3891.
Therefore, we will get 12e2−32=12(e2−3)=12(7.3891−3)=12×4.3891.\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 2.1945.2.1945.

Note: By linearity property of the integration, we will get ∫ab(px+qy)dx=p∫abxdx+q∫abydx.\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∫ab(x+y)dx=k∫abxdx+k∫abydx.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.e=2.71828.