My Math Forum split multiplication in integral

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 July 14th, 2017, 07:29 AM #1 Newbie   Joined: Jul 2017 From: Stuttgart, Germany Posts: 1 Thanks: 0 split multiplication in integral Hi, I've got an equation that looks like this: $\displaystyle G = \int_0^X 4\pi (g-1) r^2 \left( 1-\frac{3r}{4R} + \frac{3r^3}{16R^3} \right) dr$ Now I would like to separate this integral into $\displaystyle G = \int_0^X 4\pi (g-1) r^2 dr + Y$ or $\displaystyle G = \int_0^X 4\pi (g-1) r^2 dr * Y$ where Y does not contain g. Is this possible? Last edited by skipjack; July 14th, 2017 at 10:03 AM.
July 14th, 2017, 10:08 AM   #2
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 Originally Posted by Fadabi Hi, I've got an equation that looks like this: $\displaystyle G = \int_0^X 4\pi (g-1) r^2 \left( 1-\frac{3r}{4R} + \frac{3r^3}{16R^3} \right) dr$ Now I would like to separate this integral into $\displaystyle G = \int_0^X 4\pi (g-1) r^2 dr + Y$ or $\displaystyle G = \int_0^X 4\pi (g-1) r^2 dr * Y$ where Y does not contain g. Is this possible?
why not just complete the integral, it's not hard

$\displaystyle G = \int_0^X 4\pi (g-1) r^2 \left( 1-\frac{3r}{4R} + \frac{3r^3}{16R^3} \right) dr =\dfrac{\pi (g-1) X^3 \left(32 R^3-18 R^2 X+3 X^3\right)}{24 R^3}$

 July 14th, 2017, 01:08 PM #3 Global Moderator   Joined: May 2007 Posts: 6,610 Thanks: 616 Does g depend on r? If not, direct integration (romsek) works. If g depends on r, what you want is unlikely.

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