Average Value of a Function

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

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

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

Since $$$\Delta{x}=\frac{{{b}-{a}}}{{n}}$$$ then $$${n}=\frac{{{b}-{a}}}{{\Delta{x}}}$$$ and the average value becomes $$$\frac{{{f{{\left({{x}_{{1}}^{{\star}}}\right)}}}+{f{{\left({{x}_{{2}}^{{\star}}}\right)}}}+\ldots+{f{{\left({{x}_{{n}}^{{\star}}}\right)}}}}}{{\frac{{{b}-{a}}}{{\Delta{x}}}}}=$$$

$$$=\frac{{1}}{{{b}-{a}}}{\left({f{{\left({{x}_{{1}}^{{\star}}}\right)}}}+{f{{\left({{x}_{{2}}^{{\star}}}\right)}}}+\ldots+{f{{\left({{x}_{{n}}^{{\star}}}\right)}}}\right)}\Delta{x}=$$$

$$$=\frac{{1}}{{{b}-{a}}}{\sum_{{{i}={1}}}^{{n}}}{f{{\left({{x}_{{i}}^{{\star}}}\right)}}}\Delta{x}$$$.

If we let $$${n}$$$ increase, we would be computing the average value of a large number of closely spaced values. The limiting value is $$$\lim_{{{n}\to\infty}}\frac{{1}}{{{b}-{a}}}{\sum_{{{i}={1}}}^{{n}}}{f{{\left({{x}_{{i}}^{{\star}}}\right)}}}\Delta{x}=\frac{{1}}{{{b}-{a}}}{\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{d}{x}$$$ by the definition of definite integral.

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

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

Here, $$${a}={0}$$$ and $$${b}={2}$$$, therefore $$${f}_{{{a}{v}{e}}}=\frac{{1}}{{{2}-{0}}}{\int_{{0}}^{{2}}}{\left({4}-{{x}}^{{2}}\right)}{d}{x}=\frac{{1}}{{2}}{\left({4}{x}-\frac{{1}}{{3}}{{x}}^{{3}}\right)}{{\mid}_{{0}}^{{2}}}=\frac{{1}}{{2}}{\left({4}\cdot{2}-\frac{{1}}{{3}}\cdot{{2}}^{{3}}\right)}=\frac{{8}}{{3}}$$$.

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

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

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

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

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

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