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 Foundations, we have:
${\begin{array}{rcl}\displaystyle {f_{\text{avg}}}&=&\displaystyle {{\frac {1}{2-0}}\int _{0}^{2}2x^{3}(1+x^{2})^{4}~dx}\\&&\\&=&\displaystyle {\int _{0}^{2}x^{3}(1+x^{2})^{4}~dx.}\\\end{array}}$

Step 2:
Now, we use $u$-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
${\begin{array}{rcl}\displaystyle {f_{\text{avg}}}&=&\displaystyle {\int _{0}^{2}x\cdot x^{2}(1+x^{2})^{4}~dx}\\&&\\&=&\displaystyle {{\frac {1}{2}}\int _{1}^{5}(u-1)u^{4}~du}\\&&\\&=&\displaystyle {{\frac {1}{2}}\int _{1}^{5}(u^{5}-u^{4})~du.}\\\end{array}}$

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

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

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

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