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Average Value of a Function

It is easy to compute average value of finitely many values y1{y}_{{1}}, y2{y}_{{2}},...,yn{y}_{{n}}: yave=y1+y2+…+ynn{y}_{{{a}{v}{e}}}=\frac{{{y}_{{1}}+{y}_{{2}}+\ldots+{y}_{{n}}}}{{n}}.

But how to compute average of infinitely many values? In general, let's try to compute the average value of a function y=f(x){y}={f{{\left({x}\right)}}}, a≤x≤b{a}\le{x}\le{b}.

As always we start by dividing the interval [a,b]{\left[{a},{b}\right]} into n{n} equal subintervals, each with length Δx=b−an\Delta{x}=\frac{{{b}-{a}}}{{n}}. Then we choose points x1⋆{{x}_{{1}}^{{\star}}}, . . . , xn⋆{{x}_{{n}}^{{\star}}} in successive subintervals and calculate the average of the numbers f(x1⋆){f{{\left({{x}_{{1}}^{{\star}}}\right)}}}, . . . , f(xn⋆){f{{\left({{x}_{{n}}^{{\star}}}\right)}}}: f(x1⋆)+f(x2⋆)+…+f(xn⋆)n\frac{{{f{{\left({{x}_{{1}}^{{\star}}}\right)}}}+{f{{\left({{x}_{{2}}^{{\star}}}\right)}}}+\ldots+{f{{\left({{x}_{{n}}^{{\star}}}\right)}}}}}{{n}}.

Since Δx=b−an\Delta{x}=\frac{{{b}-{a}}}{{n}} then n=b−aΔx{n}=\frac{{{b}-{a}}}{{\Delta{x}}} and the average value becomes f(x1⋆)+f(x2⋆)+…+f(xn⋆)b−aΔx=\frac{{{f{{\left({{x}_{{1}}^{{\star}}}\right)}}}+{f{{\left({{x}_{{2}}^{{\star}}}\right)}}}+\ldots+{f{{\left({{x}_{{n}}^{{\star}}}\right)}}}}}{{\frac{{{b}-{a}}}{{\Delta{x}}}}}=

=1b−a(f(x1⋆)+f(x2⋆)+…+f(xn⋆))Δx==\frac{{1}}{{{b}-{a}}}{\left({f{{\left({{x}_{{1}}^{{\star}}}\right)}}}+{f{{\left({{x}_{{2}}^{{\star}}}\right)}}}+\ldots+{f{{\left({{x}_{{n}}^{{\star}}}\right)}}}\right)}\Delta{x}=

=1b−a∑i=1nf(xi⋆)Δx=\frac{{1}}{{{b}-{a}}}{\sum_{{{i}={1}}}^{{n}}}{f{{\left({{x}_{{i}}^{{\star}}}\right)}}}\Delta{x}.

If we let n{n} increase, we would be computing the average value of a large number of closely spaced values. The limiting value is lim⁡n→∞1b−a∑i=1nf(xi⋆)Δx=1b−a∫abf(x)dx\lim_{{{n}\to\infty}}\frac{{1}}{{{b}-{a}}}{\sum_{{{i}={1}}}^{{n}}}{f{{\left({{x}_{{i}}^{{\star}}}\right)}}}\Delta{x}=\frac{{1}}{{{b}-{a}}}{\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{d}{x} by the definition of definite integral.

Average Value of a Function f{f{}} on the interval [a,b]{\left[{a},{b}\right]} is fave=1b−a∫abf(x)dx{f}_{{{a}{v}{e}}}=\frac{{1}}{{{b}-{a}}}{\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{d}{x}.

For a positive function, we can think of this definition as saying areawidth=average height\frac{\text{area}}{\text{width}}=\text{average height}.

Example 1. Find the average value of function f(x)=4−x2{f{{\left({x}\right)}}}={4}-{{x}}^{{2}} on interval [0,2]{\left[{0},{2}\right]}.

Here, a=0{a}={0} and b=2{b}={2}, therefore fave=12−0∫02(4−x2)dx=12(4x−13x3)∣02=12(4⋅2−13⋅23)=83{f}_{{{a}{v}{e}}}=\frac{{1}}{{{2}-{0}}}{\int_{{0}}^{{2}}}{\left({4}-{{x}}^{{2}}\right)}{d}{x}=\frac{{1}}{{2}}{\left({4}{x}-\frac{{1}}{{3}}{{x}}^{{3}}\right)}{{\mid}_{{0}}^{{2}}}=\frac{{1}}{{2}}{\left({4}\cdot{2}-\frac{{1}}{{3}}\cdot{{2}}^{{3}}\right)}=\frac{{8}}{{3}}.

Now, recall that Mean Value Theorem for Integrals states that there exists such number c{c} from [a,b]{\left[{a},{b}\right]} that f(c)=1b−a∫abf(x)dx{f{{\left({c}\right)}}}=\frac{{1}}{{{b}-{a}}}{\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{d}{x}.

This means that there exists such number (or numbers) c{c} that f(c)=fave{f{{\left({c}\right)}}}={f}_{{{a}{v}{e}}}. In other words function takes its average value at least once.

Example 2. Find such c{c} in interval [0,2]{\left[{0},{2}\right]} that f(c)=fave{f{{\left({c}\right)}}}={f}_{{{a}{v}{e}}} for y=4−x2{y}={4}-{{x}}^{{2}} on [0,2]{\left[{0},{2}\right]}.

In Example 1 we've found that fave=83{f}_{{{a}{v}{e}}}=\frac{{8}}{{3}}.

Therefore, we need to find such c{c} that f(c)=4−c2=83{f{{\left({c}\right)}}}={4}-{{c}}^{{2}}=\frac{{8}}{{3}}.

This gives c2=4−83=43{{c}}^{{2}}={4}-\frac{{8}}{{3}}=\frac{{4}}{{3}} or c=±23{c}=\pm\frac{{2}}{\sqrt{{{3}}}}, however c=−23{c}=-\frac{{2}}{\sqrt{{{3}}}} doesn't belong to interval [0,2]{\left[{0},{2}\right]}, therefore the only applicable number is c=23{c}=\frac{{2}}{\sqrt{{{3}}}}.