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:
Since $(1+x^{2})^{2}=1+2x^{2}+x^{4}$, we have
$(1+x^{2})^{4}=(1+x^{2})^{2}(1+x^{2})^{2}=(1+2x^{2}+x^{4})(1+2x^{2}+x^{4})=1+4x^{2}+6x^{4}+4x^{6}+x^{8}$.
So, the integral becomes $f_{\text{avg}}=\int _{0}^{2}x^{3}(1+4x^{2}+6x^{4}+4x^{6}+x^{8})dx$

Step 3:
We distribute to get
$f_{\text{avg}}=\int _{0}^{2}x^{3}(1+4x^{2}+6x^{4}+4x^{6}+x^{8})dx=\int _{0}^{2}(x^{3}+4x^{5}+6x^{7}+4x^{9}+x^{11})dx$.
Now, we integrate to get
$f_{\text{avg}}=\left.{\frac {x^{4}}{4}}+4{\frac {2x^{6}}{3}}+{\frac {3x^{8}}{4}}+{\frac {2x^{10}}{5}}+{\frac {x^{12}}{12}}\right|_{0}^{2}$

Final Answer:

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