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.
$\displaystyle h(x)=\int_{2}^{x^2}\sqrt{u-1}du$
Change of variable. Let y = u - 1. ThenOriginally Posted by Zeppelin
$\displaystyle h(x) = \int_1^{x^2-1} dy\, \sqrt{y}$
$\displaystyle = \frac{2}{3}y^{3/2}|_1^{x^2-1}$
$\displaystyle = \frac{2}{3}(x^2-1)^{3/2} - \frac{2}{3}(1)^{3/2}$
$\displaystyle = \frac{2}{3}(x^2-1)^{3/2} - \frac{2}{3}$
-Dan
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).Originally Posted by Zeppelin
If you had,
$\displaystyle f(x)=\int_2^x \sqrt{u-1}du$
Then,
$\displaystyle f'(x)=\sqrt{x-1}$
But they have a square.
You can image $\displaystyle h(x)$ as a composition of functions.
$\displaystyle f(g(x))$ where $\displaystyle g(x)=x^2$.
Then its derivative involves the chain rule.
$\displaystyle g'(x)f'(g(x))$
But,
$\displaystyle g'(x)=2x$
And,
$\displaystyle f'(x)=\sqrt{x-1}$ thus $\displaystyle f'(g(x))=\sqrt{x^2-1}$
Thus, the derivative is,
$\displaystyle 2x\sqrt{x^2-1}$