Looking at some of the older stuff and for ∫1/10(x)(6x^2-15) dx became 1/120 ∫12x(6x^2-15) dx, how was the 1/120 ∫12x arrived at?

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- February 22nd 2013, 01:20 PMlandmarkQuestion
Looking at some of the older stuff and for ∫1/10(x)(6x^2-15) dx became 1/120 ∫12x(6x^2-15) dx, how was the 1/120 ∫12x arrived at?

- February 22nd 2013, 01:43 PMcoffee2000Re: Question
1/10 = 1/120 * 12

and we know that you can move a constant outside of the integral - February 22nd 2013, 01:47 PMlandmarkRe: Question
I can see that but why 12?

- February 22nd 2013, 01:49 PMcoffee2000Re: Question
probably to make u-substitution more straightforward

i.e.

u=6x^2-15

du=12x dx - February 22nd 2013, 02:03 PMlandmarkRe: Question
Not sure if I quite get it but thanks

- February 22nd 2013, 02:16 PMcoffee2000Re: Question
EDIT:

**1/120 ∫12x(6x^2-15) dx**

=1/120**∫ 72x^3-180x dx**I'm sorry i forgot that it was possible that you didn't cover u-substitution, which is just a method of integration and all you had to do was multiply it out and integrate the polynomial! sorry! :)

and 1/120 was probably to avoid fractions during integration - February 22nd 2013, 11:55 PMlandmarkRe: Question
Thanks for that I have to get used to such manipulation, I have never been on one of these forums before and can not figure out how to do a smiley face

- February 23rd 2013, 06:43 AMtopsquarkRe: Question