# 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
 HW Helper Sci Advisor Thanks P: 24,423 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).
 HW Helper Sci Advisor P: 1,276 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
 HW Helper Sci Advisor Thanks P: 24,423 What did you put into the integrator??
 P: 25 (2sinx)^2*sqrt[4-(2sinx)^2]
HW Helper
Sci Advisor
Thanks
P: 24,423
 Quote by LHC (2sinx)^2*sqrt[4-(2sinx)^2]
You forgot the part of the integrand that's coming from the dx in your substitution.
 P: 25 Alright, I shall try to fix it. (Gotta go, it's time for dinner.) Thanks for your help!

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