My Math Forum Integration

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July 9th, 2017, 03:44 AM   #1
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Integration

Can any show me the steps that were followed to the answer for the attached integration problem?
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Last edited by skipjack; July 9th, 2017 at 04:52 AM.

 July 9th, 2017, 04:05 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,109 Thanks: 855 Pretty basic isn't it? I presume you know that $\displaystyle \int C f(x)dx= C\int f(x)dx$ so you can just take that "$\displaystyle \frac{9V^2dx}{8Gbh^6}$" outside the integral and multiply by it after doing the integration. That "dx" in the numerator looks peculiar, but the whole integration is equal to "dU" so I presume there will be another integration with respect to x. Also $\displaystyle \int f(x)+ g(x) dx= \int f(x)dx+ \int g(x) dx$ so you can do each of $\displaystyle \int h^4 dy$, $\displaystyle \int -8h^2y^2 dy$, $\displaystyle \int 16y^4dy$ separately. Of course, since, as before $\displaystyle \int C f(x)dx= C\int f(x)dx$, we can write those as $\displaystyle h^4\int dy= h^4\int y^0 dy$, $\displaystyle -8h^2\int y^2 dy$, and $\displaystyle 16\int y^4 dy$. The last "step" is to use the fact that $\displaystyle \int y^n dy= \frac{1}{n+1}y^{n+1}dy$ so that $\displaystyle \int_{-h/2}^{h/2} y^n dy= \frac{1}{n+1}\left((h/2)^{n+1}- (-h/2)^{n+1}\right)$. Thanks from hatchelhoff Last edited by skipjack; July 9th, 2017 at 04:49 AM.

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