Hello

I have to find the integer of (1/10)*x(6x^2-15)

my attempt

(1/10)*x=(x/10)*x=x^2/20

and then I get a bit stuck , do I square (6x^2-15)

Or am supposed to use one of the rules

Just can't get my head around

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- Feb 15th 2013, 01:52 AMduggielangerStumped how to do this integration
Hello

I have to find the integer of (1/10)*x(6x^2-15)

my attempt

(1/10)*x=(x/10)*x=x^2/20

and then I get a bit stuck , do I square (6x^2-15)

Or am supposed to use one of the rules

Just can't get my head around - Feb 15th 2013, 02:54 AMa tutorRe: Stumped how to do this integration
You can multiply it out to get $\displaystyle \frac{6x^3}{10}-\frac{15x}{10}=\frac{3x^3}{5}-\frac{3x}{2}$ and then integrate.

- Feb 15th 2013, 03:36 AMduggielangerRe: Stumped how to do this integration
Ah yes of course I get in now thank you very much I think I can work it out from here

Thank you - Feb 15th 2013, 06:45 AMProve ItRe: Stumped how to do this integration
An alternative to multiplying everything out:

$\displaystyle \displaystyle \begin{align*} \int{ \frac{1}{10} x \left( 6x^2 - 15 \right) dx } &= \frac{1}{120} \int{ 12x \left( 6x^2 - 15 \right) dx } \end{align*}$

Now let $\displaystyle \displaystyle \begin{align*} u = 6x^2 - 15 \implies du = 12x \, dx \end{align*}$ and the integral becomes

$\displaystyle \displaystyle \begin{align*} \frac{1}{120} \int{ 12x \left( 6x^2 - 15 \right) dx } &= \frac{1}{120} \int{ u\, du } \\ &= \frac{1}{120} \left( \frac{u^2}{2} \right) + C \\ &= \frac{1}{240} u^2 + C \\ &= \frac{1}{240} \left( 6x^2 - 15 \right) ^2 + C \end{align*}$ - Feb 15th 2013, 09:26 AMduggielangerRe: Stumped how to do this integration
Sorry getting confused now , I will show you the whole question and my answer

evaluate the function (sorry don't know how to put the intergral sign in) but its 6 at the top and 3 at the bottom of (1/10)*x(6x^2-15)

so

(1/10)*x(6x^2-15)=(3x^3/5)-(3x/2)

Now to integrate it to get (3x^4/20)-(3x^2/4)

Now to add 6 for x and the 3 for x

((3*6^4)/20))-((3*6^2)/4))=167.4

((3*3^4)/20))-((3*3^2)/4))=5.4

and finally 167.4-5.4=162

I don't get the way you've done it prove it is the composite rule - Feb 15th 2013, 06:29 PMhollywoodRe: Stumped how to do this integration
If you are just learning integration, you will soon get to the substitution method, which is what Prove It used. The method of multiplying it out and using the rule for powers of x works, too, and should give the same answer:

$\displaystyle \frac{1}{240} \left( 6x^2 - 15 \right) ^2 =$

$\displaystyle \frac{1}{240} \left( 36x^4 - 180x^2 + 225 \right) =$

$\displaystyle \frac{36}{240}x^4 - \frac{180}{240}x^2 + \frac{225}{240} =$

$\displaystyle \frac{3x^4}{20} - \frac{3x^2}{4} + \frac{15}{16}$

which is the same as the result you got, $\displaystyle \frac{3x^4}{20} - \frac{3x^2}{4}$, except for a constant. (Remember that "+C" that you're supposed to add on when you're finished).

- Hollywood - Feb 16th 2013, 02:25 AMduggielangerRe: Stumped how to do this integration
I am just learning integration and yes we haven't done that way yet , thank you everyone for your help

- Feb 22nd 2013, 12:50 PMlandmarkRe: Stumped how to do this integration
Just wondering where you got 1/120 ∫12x