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I am working on a practice problem for an exam, and I am having trouble with this particular problem. How would I go about finding the average value of $x^2\sqrt{x+1}$? I plugged it into the average value function and got the following:

$\frac{1}{3}$$\int_{0}^{3} x^2\sqrt{x+1} dx$

But now I'm stumped on how to proceed. I tried u substitution but can't seem to figure it out. I looked at wolfram alpha and they used $\sqrt{x+1}$ as u. That doesn't seem to work for me. How did they make that work?

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$$ \int_0^3 x^2 \sqrt{x+1} dx = \int_{u=1}^2 (u^2-1)^2 u ( 2u \, du ) $$ You can take it from there.

I suspect you problem was not getting the limits of integration right; $u$ does not go from 0 to 3.

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For your integral: $$\frac 13 \int_{0}^{3} x^2\sqrt{x+1} dx$$ $$ u = \sqrt{x+1} \implies du = \frac{1}{2\sqrt{x+1}}\,dx \iff 2udu = dx$$ and $u^2 = x+1\implies x = u^2-1$.

At $x=0, u = \sqrt{0+1} = 1$ and at $x=3, u = \sqrt{3+1}=2$.

That gives us $$\frac 13 \int_1^2 (u^2 - 1)^2(u)(2udu) = 2\int u^2(u^4 - 2u^2 - 1)\,du$$

Can you take it from here?

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