# Definite Integral

by LHC
Tags: definite, integral
 P: 25 The question in my textbook was: $$\int_{0}^{2} x^2 \sqrt{4-x^2} dx$$ I decided to just leave out the lower and upper limits for now, and just solve $$\int x^2 \sqrt{4-x^2} dx$$. (It's a bit long, but I assure you I did the work.) Upon making the substitution of $$x = 2 \sin \theta$$, I got it down to: $$\int x^2 \sqrt{4-x^2} dx = \frac{8}{3} \sin^3 \theta + C$$ Now, I'm transforming it back in terms of x, so $$\sin \theta = \frac{x}{2}$$ So, I thought it would make this whole thing: $$\int x^2 \sqrt{4-x^2} dx = \frac{8}{3} \times \frac{x^3}{8}$$, so $$\int x^2 \sqrt{4-x^2} dx = \frac{x^3}{3}$$ If I finish up the problem by using the limits of 0 and 2, I get: $$\int_{0}^{2} x^2 \sqrt{4-x^2} dx = \frac{8}{3}$$ But the answer I got from the calculator was around 3.14 (not pi, though). Could someone please tell me where I went wrong? Thanks
 Sci Advisor HW Helper Thanks P: 25,235 Well, something went wrong in the long part you aren't telling us about. Because the derivative of x^3/3 is DEFINITELY not x^2*sqrt(4-x^2).
 Sci Advisor HW Helper P: 1,275 Also, FYI the integral should be exactly pi (so says maple).
 P: 25 Definite Integral Well, I think it's some misunderstanding on my part from the conversion between x and theta. I used the Mathematica online integrator to verify and its answer was: $$\frac{8}{3} \sqrt{\cos^2 x} \sin^2 x \tan x$$...which, when I simplify (and perhaps this is the part where I'm wrong), I just end up with $$\frac{8}{3} \sin^3 \theta$$... which is where I started off =P
 Sci Advisor HW Helper Thanks P: 25,235 What did you put into the integrator??
 P: 25 (2sinx)^2*sqrt[4-(2sinx)^2]