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[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