# first fundamental theorem of calc

• Jan 30th 2007, 05:56 PM
thedoge
first fundamental theorem of calc
f(x)=integral[4,x^2] (1/4t^2-1)^15 dt

the only thing tripping me up here is the ^15.

i accidentally did the problem through without the ^15, but dont know where i should have used it anyway

i set x^2=u

so

=(1/4*u^2-1)^15(2x)
=(1/4*(x^2)^2-1)^15(2x)
=(1/4*x^4-1)^15(2x)

do i continue from here or did i make a mistake?
• Jan 30th 2007, 10:11 PM
earboth
Quote:

Originally Posted by thedoge
f(x)=integral[4,x^2] (1/4t^2-1)^15 dt...

Hello,

$\displaystyle \int_{4}^{x^2}{\left( \frac{1}{4}t^2-1 \right)^{15} dt}$

The only way I know to do this problem is expand the bracket (but slavery is still prohibited in the US, isn't it?). So I guess that there is a typo.

If you want I can post the result but the numbers are really enormous.

EB
• Jan 31st 2007, 03:59 AM
ThePerfectHacker
Quote:

Originally Posted by earboth
Hello,

$\displaystyle \int_{4}^{x^2}{\left( \frac{1}{4}t^2-1 \right)^{15} dt}$