Hi. I missed today's class and was wondering if someone could assist me with solving this problem.
Thanks!
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P.S. I attached the picture of the problem so that noone gets confused.
Okay, I've lost you. I see that you've brought a 1/4 to the outside by squaring 4x^2, but how come the range of the integral also changes?
Also, how do I use sin$\displaystyle \theta$ to solve the rest of the problem?
I don't mean for you to simply solve the problem for me; I simply want to follow how you're getting to that point.
Thanks so much.
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x varies from -1/4 to +1/4
Therefore u=4x varies from -1 to+1
And du = 4 dx => dx = du / 4 (this is where 1/4 comes from)
$\displaystyle u = \sin \theta$ and u varies from -1 to 1 therefore $\displaystyle \theta$ varies from $\displaystyle -\frac{\pi}{2}$ to $\displaystyle +\frac{\pi}{2}$
And $\displaystyle du = \cos \theta d\theta$