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:

\(\displaystyle{f}_{{{a}{v}{g}}}={\frac{{{1}}}{{{6}}}}{\int_{{-{3}}}^{{{3}}}}{\frac{{{x}^{{{2}}}}}{{{\left({x}^{{{3}}}+{30}\right)}^{{{2}}}}}}{\left.{d}{x}\right.}={\frac{{{1}}}{{{57}}}}\)

\(\displaystyle{f}_{{{a}{v}{g}}}={\frac{{{1}}}{{{b}-{a}}}}{\int_{{{a}}}^{{{b}}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}\)

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:

\(\displaystyle{f}_{{{a}{v}{g}}}={\frac{{{1}}}{{{6}}}}{\int_{{-{3}}}^{{{3}}}}{\frac{{{x}^{{{2}}}}}{{{\left({x}^{{{3}}}+{30}\right)}^{{{2}}}}}}{\left.{d}{x}\right.}={\frac{{{1}}}{{{57}}}}\)