Hi. I missed today's class and was wondering if someone could assist me with solving this problem.

Thanks!

-star

P.S. I attached the picture of the problem so that noone gets confused.

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- Mar 17th 2010, 01:23 PMstarblessintegral between 1/4 and -1/4 of (1/(sqrt(1-(4x)^2))dx
Hi. I missed today's class and was wondering if someone could assist me with solving this problem.

Thanks!

-star

P.S. I attached the picture of the problem so that noone gets confused. - Mar 17th 2010, 01:28 PMrunning-gag
Hi

Through $\displaystyle u = 4x$

$\displaystyle \int_{-\frac14}^{\frac14} \frac{dx}{\sqrt{1-(4x)^2}} = \frac14 \int_{-1}^{1} \frac{du}{\sqrt{1-u^2}}$

Then you can use $\displaystyle u = \sin \theta$ - Mar 17th 2010, 01:37 PMstarbless
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.

-star - Mar 17th 2010, 01:49 PMrunning-gag
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 $\displaystyle du = \cos \theta d\theta$ - Mar 17th 2010, 02:24 PMGeneral
**You should notice that this is an improper integral, it should be solved in terms of limits ..**