Stumped how to do this integration

**duggielanger** · February 15th 2013, 01:52 AM

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

**a tutor** · February 15th 2013, 02:54 AM

You can multiply it out to get $\frac{6x^3}{10}-\frac{15x}{10}=\frac{3x^3}{5}-\frac{3x}{2}$ and then integrate.

**duggielanger** · February 15th 2013, 03:36 AM

Ah yes of course I get in now thank you very much I think I can work it out from here
Thank you

**Prove It** · February 15th 2013, 06:45 AM

Originally Posted by duggielanger

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

An alternative to multiplying everything out:

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

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

$\displaystyle \begin{align*} \frac{1}{120} \int{ 12x \left( 6x^2 - 15 \right) dx } &= \frac{1}{120} \int{ u\, du } \\ &= \frac{1}{120} \left( \frac{u^2}{2} \right) + C \\ &= \frac{1}{240} u^2 + C \\ &= \frac{1}{240} \left( 6x^2 - 15 \right) ^2 + C \end{align*}$

**duggielanger** · February 15th 2013, 09:26 AM

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

**hollywood** · February 15th 2013, 06:29 PM

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:

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

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

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

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

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

- Hollywood

**duggielanger** · February 16th 2013, 02:25 AM

I am just learning integration and yes we haven't done that way yet , thank you everyone for your help

**landmark** · February 22nd 2013, 12:50 PM

Just wondering where you got 1/120 ∫12x

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