What is the average value of $y=x^2\sqrt{x^3+1}$ on the interval $[0,2]$ ?

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The average value of $y$ is given by:

$$\frac{1}{b-a}\int\limits_a^b f(x)   dx$$ $$= \frac{1}{2-0}\int\limits_0^2 x^2\sqrt{x^3+1}   dx$$

We can integrate using the substitution $u=x^3+1$.

$$u = x^3+1 \hskip{3em} du = 3x^2   dx \hskip{3em} dx = \frac{du}{3x^2}$$

Substituting these values into our integral:

$$\frac{1}{2}\int\limits_1^9 x^2\sqrt{u}   \frac{du}{3x^2}$$ $$= \frac{1}{6} \int\limits_1^9 \sqrt{u}   du$$ $$= \frac{1}{6} \Big[ \frac{2}{3}u^{\frac{3}{2}} \Big]_1^9$$ $$= \frac{1}{9}\left(9^{\frac{3}{2}}-1^{\frac{3}{2}}\right)$$ $$= \frac{1}{9}(27-1)$$ $$= \boxed{\frac{26}{9}}$$