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
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{4948}{5}}

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