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integral calculus

Posted October 31st, 2009 by deepak
in

i am not able to solve this basic calculus equation
$ \int_{1}^{3}[x]cos(\frac{\pi}{2}(x-[x]))dx $

‹ Trigonometric Limits nth-order PDE ›

Comments

Where

On October 31st, 2009 deepak says:

Here [] represents greatest integer function

answer

On October 31st, 2009 Structure says:

It is very easy to solve your problem.
All you have to do is to understand some few details.

$ \int_{1}^{3}[x]\cos\frac{\pi}{2}(x-[x])dx = \int_{1}^{2}[x]cos\frac{\pi}{2}(x-[x])dx + \int_{2}^{3}[x]cos\frac{\pi}{2}(x-[x])dx= \int_{1}^{2}cos\frac{\pi}{2}(x-1)dx + \int_{2}^{3}2cos\frac{\pi}{2}(x-2)dx = \int_{0}^{1}\cos\frac{\pi}{2}(x)dx+ \int_{0}^{1}2cos\frac{\pi}{2}xdx =3 \int_{0}^{1}cos\frac{\pi}{2}xdx=3\frac{2}{\pi} $
I have used change of variable theorem x-1=y and x-2=y

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