# Thread: Stumped how to do this integration

1. ## Stumped 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

2. ## Re: 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.

3. ## Re: 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

4. ## Re: Stumped how to do this integration

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 \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*}

5. ## Re: 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

6. ## Re: 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

7. ## Re: 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

8. ## Re: Stumped how to do this integration

Just wondering where you got 1/120 ∫12x