The question in my textbook was: [tex]\int_{0}^{2} x^2 \sqrt{4-x^2} dx[/tex] I decided to just leave out the lower and upper limits for now, and just solve [tex]\int x^2 \sqrt{4-x^2} dx[/tex]. (It's a bit long, but I assure you I did the work.) Upon making the substitution of [tex]x = 2 \sin \theta[/tex], I got it down to: [tex]\int x^2 \sqrt{4-x^2} dx = \frac{8}{3} \sin^3 \theta + C[/tex] Now, I'm transforming it back in terms of x, so [tex]\sin \theta = \frac{x}{2}[/tex] So, I thought it would make this whole thing: [tex]\int x^2 \sqrt{4-x^2} dx = \frac{8}{3} \times \frac{x^3}{8}[/tex], so [tex]\int x^2 \sqrt{4-x^2} dx = \frac{x^3}{3}[/tex] If I finish up the problem by using the limits of 0 and 2, I get: [tex]\int_{0}^{2} x^2 \sqrt{4-x^2} dx = \frac{8}{3}[/tex] 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
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).
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: [tex]\frac{8}{3} \sqrt{\cos^2 x} \sin^2 x \tan x[/tex]...which, when I simplify (and perhaps this is the part where I'm wrong), I just end up with [tex]\frac{8}{3} \sin^3 \theta [/tex]... which is where I started off =P