Question

Find the average value of the function \({\rm{f(x) = tsin}}\left( {{{\rm{t}}^{\rm{2}}}} \right)\) on the interval \({\rm{(0,2)}}{\rm{.}}\)

Short answer

The resulted value of the function is \({\rm{0}}{\rm{.0069}}{\rm{.}}\)

Step-by-step solution

Defining average of the function

The integral of a function divided by the length of the interval is the average value of the function over the interval.

\({\rm{y = f(x)}}\) is the average value of the function on the interval \({\rm{(a, b),}}\)

\({{\rm{f}}_{{\rm{avg}}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{b - a}}}}\int_{\rm{a}}^{\rm{b}} {\rm{f}} {\rm{(x)dx}}\)

Estimating the average value of the function

The average value of the function\({\rm{f(x) = tsin}}\left( {{{\rm{t}}^{\rm{2}}}} \right)\)on the interval\({\rm{(0,10)}}\)is

\(\begin{array}{*{20}{c}}{{{\rm{f}}_{{\rm{avg}}}}}&{{\rm{ = }}\frac{{\rm{1}}}{{{\rm{10 - 0}}}}\int_{\rm{0}}^{{\rm{10}}} {\rm{t}} {\rm{sin}}\left( {{{\rm{t}}^{\rm{2}}}} \right){\rm{dt}}}\\{{{\rm{f}}_{{\rm{avg}}}}}&{{\rm{ = }}\frac{{\rm{1}}}{{{\rm{20}}}}\int_{\rm{0}}^{{\rm{10}}} {{\rm{sin}}} \left( {{{\rm{t}}^{\rm{2}}}} \right){\rm{(2tdt)}}}\end{array}\)

By substituting\({{\rm{t}}^{\rm{2}}}{\rm{ = u}}\)and\({\rm{2 t d t = d u}}\),

Therefore, the average of the function is \({\rm{0}}{\rm{.0069}}{\rm{.}}\)