
#1
Jul2208, 04:45 PM

P: 25

The question in my textbook was:
[tex]\int_{0}^{2} x^2 \sqrt{4x^2} dx[/tex] I decided to just leave out the lower and upper limits for now, and just solve [tex]\int x^2 \sqrt{4x^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{4x^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{4x^2} dx = \frac{8}{3} \times \frac{x^3}{8}[/tex], so [tex]\int x^2 \sqrt{4x^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{4x^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 



#2
Jul2208, 05:00 PM

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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(4x^2).




#3
Jul2208, 05:01 PM

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




#4
Jul2208, 05:07 PM

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: [tex]\frac{8}{3} \sqrt{\cos^2 x} \sin^2 x \tan x[/tex]...which, when I simplify (and perhaps this is the part where I'm wrong), I just end up with [tex]\frac{8}{3} \sin^3 \theta [/tex]... which is where I started off =P 



#5
Jul2208, 05:16 PM

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What did you put into the integrator??




#6
Jul2208, 05:19 PM

P: 25

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




#7
Jul2208, 05:24 PM

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P: 25,160





#8
Jul2208, 05:29 PM

P: 25

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



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