This is a problem that someone presented to me and I tried to do. My work is shown below the problem. I got my answer (or I thought I did), but it is not in agreement with my TI-84's average value of $\displaystyle \frac{1}{3-1} \int_1^3 x^2\sqrt{3x+1} \, dx = 12.135237$. My answer below is different, and I can't seem to find the error in my work.
The average value of a function $f$ over the interval $[a,b]$ is given by $$\text{Average value}=\frac 1{b-a} \int_a^b f(x) \, dx.$$ What is the average value of $f(x)=x^2\sqrt{3x+1}$ over the interval $[1,3]$?
Let $u=3x+1$. Then $\displaystyle \frac {du}3= dx \text{ and } x^2=\frac{u^2-2u+1}9$. \begin{align*} \frac 1{3-1} \int_1^3 x^2 \sqrt{3x+1} \, dx &= \frac 12 \int_4^{10} \frac{u^2-2u+1}9 \sqrt{u} \, \frac {du}3 \\ &= \frac 12 \cdot \frac 19 \cdot \frac 13 \int_4^{10} u^{5/2} - 2u^{3/2} + u^{1/2} \, du \\ &= \frac 1{54} \left(\frac 27 u^{7/2} - \frac 45 u^{5/2} + \frac 23 u^{3/2} \right) \Big\vert_4^{10} \\ &= \boxed{\frac{2230}{567} \sqrt{10} - \frac{488}{405}} \approx \boxed{33.697} \end{align*}
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$\begingroup$Only the last line of your numerical application has a mistake when evaluating the contribution from the term in $u = 4$. Indeed, we have
$$ \frac{1}{54} \left[\frac{2}{7} u^{7/2} - \frac{4}{5} u^{5/2} + \frac{2}{3} u^{3/2} \right]_{4}^{10} = \frac{2230}{567} \sqrt{10} - \frac{856}{2835} \simeq 12.1352 $$
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