# Find the average value fave of the function ff on

Find the average value fave of the function ff on the given interval.
$$\displaystyle{f{{\left({x}\right)}}}={\frac{{{x}^{{{2}}}}}{{{\left({x}^{{{3}}}+{30}\right)}^{{{2}}}}}},\ \ \ {x}\in{\left[-{3},{3}\right]}$$

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Clara Reese
The average value of the function f on the interval [a,b] is equal to:
$$\displaystyle{f}_{{{a}{v}{g}}}={\frac{{{1}}}{{{b}-{a}}}}{\int_{{{a}}}^{{{b}}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$$
The given function is
$$\displaystyle{f{{\left({x}\right)}}}={\frac{{{x}^{{{2}}}}}{{{\left({x}^{{{3}}}+{30}\right)}^{{{2}}}}}},\ \ \ {x}\in{\left[-{3},{3}\right]}$$
Thus, the average value of f is given by
$$\displaystyle{f}_{{{a}{v}{g}}}={\frac{{{1}}}{{{3}-{\left(-{3}\right)}}}}{\int_{{-{3}}}^{{{3}}}}{\frac{{{x}^{{{2}}}}}{{{\left({x}^{{{3}}}+{30}\right)}^{{{2}}}}}}{\left.{d}{x}\right.}={\frac{{{1}}}{{{6}}}}{\int_{{-{3}}}^{{{3}}}}{\frac{{{x}^{{{2}}}}}{{{\left({x}^{{{3}}}+{30}\right)}^{{{2}}}}}}{\left.{d}{x}\right.}$$
Now first calculate the indefinite integral $$\displaystyle\int{\frac{{{x}^{{{2}}}}}{{{\left({x}^{{{3}}}+{30}\right)}^{{{2}}}}}}{\left.{d}{x}\right.}$$
$$\displaystyle{\int_{{-{3}}}^{{{3}}}}{\frac{{{x}^{{{2}}}}}{{{\left({x}^{{{3}}}+{30}\right)}^{{{2}}}}}}{\left.{d}{x}\right.}=\int{\frac{{{1}}}{{{3}{u}^{{{2}}}}}}{d}{u}$$ [Apply u - substitution: $$\displaystyle{u}={x}^{{{3}}}+{30}$$]
$$\displaystyle={\frac{{{1}}}{{{3}}}}\cdot{\frac{{{u}^{{-{2}+{1}}}}}{{-{2}+{1}}}}$$
$$\displaystyle=-{\frac{{{1}}}{{{3}{u}}}}$$
$$\displaystyle=-{\frac{{{1}}}{{{3}{\left({x}^{{{3}}}+{30}\right)}}}}$$ [Substitute back: $$\displaystyle{u}={x}^{{{3}}}+{30}$$]
Therefore we get
$$\displaystyle{f}_{{{a}{v}{g}}}={\frac{{{1}}}{{{6}}}}{\int_{{-{3}}}^{{{3}}}}{\frac{{{x}^{{{2}}}}}{{{\left({x}^{{{3}}}+{30}\right)}^{{{2}}}}}}{\left.{d}{x}\right.}$$
$$\displaystyle={\frac{{{1}}}{{{6}}}}{\left({\left[-{\frac{{{1}}}{{{3}{\left({x}^{{{3}}}+{30}\right)}}}}\right]}_{{{x}={3}}}-{\left[-{\frac{{{1}}}{{{3}{\left({x}^{{{3}}}+{30}\right)}}}}\right]}_{{{x}={3}}}\right)}$$
$$\displaystyle={\frac{{{1}}}{{{6}}}}{\left[-{\frac{{{1}}}{{{171}}}}-{\left(-{\frac{{{1}}}{{{9}}}}\right)}\right]}$$
$$\displaystyle={\frac{{{1}}}{{{6}}}}\cdot{\frac{{{2}}}{{{19}}}}$$
$$\displaystyle={\frac{{{1}}}{{{57}}}}$$
Result:
$$\displaystyle{f}_{{{a}{v}{g}}}={\frac{{{1}}}{{{6}}}}{\int_{{-{3}}}^{{{3}}}}{\frac{{{x}^{{{2}}}}}{{{\left({x}^{{{3}}}+{30}\right)}^{{{2}}}}}}{\left.{d}{x}\right.}={\frac{{{1}}}{{{57}}}}$$