Solving Definite Integral: \int x^2 \sqrt{4-x^2} dx | Textbook Question Answered

[LHC]

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

 

[Dick]

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).

 

[nicksauce]

Also, FYI the integral should be exactly pi (so says maple).

 

[LHC]

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

 

[Dick]

What did you put into the integrator??

 

[LHC]

(2sinx)^2*sqrt[4-(2sinx)^2]

 

[Dick]

(2sinx)^2*sqrt[4-(2sinx)^2]

You forgot the part of the integrand that's coming from the dx in your substitution.

 

[LHC]

Alright, I shall try to fix it. (Gotta go, it's time for dinner.) Thanks for your help!