- #1
LHC
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- 0
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
[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