# What is the average value of the function g(x)=[x^{2}\sqrt{1=x^{3}}] on

Izabelle Walsh 2022-04-18 Answered
What is the average value of the function $g\left(x\right)=\left[{x}^{2}\sqrt{1={x}^{3}}\right]$ on the interval [0,2]?
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If f(x) is a continuous function on the interval [a,b], then the average value is
$\stackrel{―}{x}=\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)dx$
Here, g(x) is continuous on the interval [0,2]
Therefore, the average value is
$\stackrel{―}{x}=\frac{1}{2-0}{\int }_{0}^{2}{x}^{2}+\sqrt{1+{x}^{3}}dx$
First, compute the indefinite integral
$I=\int {x}^{2}\sqrt{1={x}^{3}}dx$
Let $u=1+{x}^{3},⇒,du=3{x}^{2}dx$
Therefore,
$I=\frac{1}{3}\int {u}^{\frac{1}{2}}du=\frac{1}{3}\frac{{u}^{\frac{3}{2}}}{\frac{3}{2}}$
$=\frac{2}{9}{\left(1={x}^{3}\right)}^{\frac{3}{2}}+C$
And finally
$\stackrel{―}{x}=\frac{1}{2}\cdot \frac{2}{9}{\left[{\left(1+{x}^{3}\right)}^{\frac{3}{2}}\right]}_{0}^{2}$
$=\frac{1}{9}\cdot \left(\left(27\right)-\left(1\right)\right)$
$=\frac{26}{9}$