# Thread: Fundamental Theorem of Calculus Help Needed

1. ## Fundamental Theorem of Calculus Help Needed

Hello all, I am new here, and would like some help please. I did look around at LaTex first in an effort to make my questions more readable.

$h(x)=\int_{2}^{x^2}\sqrt{u-1}du$

2. Originally Posted by Zeppelin
Hello all, I amnew here, and would like some help please. I did look around at LaTex first in an effort to make my questions more readable.

$h(x)=\int_{2}^{x^2}\sqrt{u-1}du$
Change of variable. Let y = u - 1. Then
$h(x) = \int_1^{x^2-1} dy\, \sqrt{y}$

$= \frac{2}{3}y^{3/2}|_1^{x^2-1}$

$= \frac{2}{3}(x^2-1)^{3/2} - \frac{2}{3}(1)^{3/2}$

$= \frac{2}{3}(x^2-1)^{3/2} - \frac{2}{3}$

-Dan

3. Originally Posted by Zeppelin
Hello all, I am new here, and would like some help please. I did look around at LaTex first in an effort to make my questions more readable.

$h(x)=\int_{2}^{x^2}\sqrt{u-1}du$
I think he means he want its derivative (it makes more sense like that because many questions want you to find the derivative of the integral invloving the fundamental theorem).

$f(x)=\int_2^x \sqrt{u-1}du$
Then,
$f'(x)=\sqrt{x-1}$
But they have a square.

You can image $h(x)$ as a composition of functions.
$f(g(x))$ where $g(x)=x^2$.
Then its derivative involves the chain rule.

$g'(x)f'(g(x))$
But,
$g'(x)=2x$
And,
$f'(x)=\sqrt{x-1}$ thus $f'(g(x))=\sqrt{x^2-1}$
Thus, the derivative is,
$2x\sqrt{x^2-1}$

4. Yes thank you Perfect Hacker, I apologize, I didn't add the question, just the equation. I was looking for the derivitive, thanks for the tutoring. I'm sure I'll be back!