009B Sample Midterm 1, Problem 2

Find the average value of the function on the given interval.

$f(x)=2x^{3}(1+x^{2})^{4},~~~[0,2]$

Foundations:
The average value of a function $f(x)$ on an interval $[a,b]$ is given by $f_{\text{avg}}={\frac {1}{b-a}}\int _{a}^{b}f(x)dx$.

Solution:

Step 1:
Using the formula given in the Foundations sections, we have:
$f_{\text{avg}}={\frac {1}{2-0}}\int _{0}^{2}2x^{3}(1+x^{2})^{4}dx=\int _{0}^{2}x^{3}(1+x^{2})^{4}dx$

Step 2:
Now, we use substitution. Let $u=1+x^{2}$. Then, $du=2xdx$ and ${\frac {du}{2}}=xdx$. Also, $x^{2}=u-1$.
We need to change the bounds on the integral. We have $u_{1}=1+0^{2}=1$ and $u_{2}=1+2^{2}=5$.
So, the integral becomes $f_{\text{avg}}=\int _{0}^{2}xx^{2}(1+x^{2})^{4}dx={\frac {1}{2}}\int _{1}^{5}(u-1)u^{4}du={\frac {1}{2}}\int _{1}^{5}(u^{5}-u^{4})du$.

Step 3:
We integrate to get
$f_{\text{avg}}=\left.{\frac {u^{6}}{12}}-{\frac {u^{5}}{10}}\right|_{1}^{5}=\left.u^{5}{\bigg (}{\frac {u}{12}}-{\frac {1}{10}}{\bigg )}\right|_{1}^{5}$

Step 4:
We evaluate to get
$f_{\text{avg}}=5^{5}{\bigg (}{\frac {5}{12}}-{\frac {1}{10}}{\bigg )}-1^{5}{\bigg (}{\frac {1}{12}}-{\frac {1}{10}}{\bigg )}=3125{\bigg (}{\frac {19}{60}}{\bigg )}-{\frac {-1}{60}}={\frac {59376}{60}}={\frac {4948}{5}}$.

Final Answer:
${\frac {4948}{5}}$

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