# 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.

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

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

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 ${c}=\frac{{2}}{\sqrt{{{3}}}}$.