# Finding average value of the function

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}$$\int_{0}^{3} x^2\sqrt{x+1} dx But now I'm stumped on how to proceed. I tried u substitution but can't seem to figure it out. I looked at wolfram alpha and they used \sqrt{x+1} as u. That doesn't seem to work for me. How did they make that work? ## 2 Answers$$ \int_0^3 x^2 \sqrt{x+1} dx = \int_{u=1}^2 (u^2-1)^2 u ( 2u \, du ) $$You can take it from there. I suspect you problem was not getting the limits of integration right; u does not go from 0 to 3. For your integral:$$\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?