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