## Definite Integral

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

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 Recognitions: Homework Help Science Advisor 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).
 Recognitions: Homework Help Science Advisor Also, FYI the integral should be exactly pi (so says maple).

## 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

 Recognitions: Homework Help Science Advisor What did you put into the integrator??
 (2sinx)^2*sqrt[4-(2sinx)^2]

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