# Thread: Solving derivative and integral using calculus fundamental theorem

1. ## Solving derivative and integral using calculus fundamental theorem

The statement says:using the calculus fundamental theorem find:
$\displaystyle\frac{d}{dx}(\displaystyle\int_{2}^{x }\displaystyle\frac{t^{3/2}}{\sqrt[ ]{t^2+17}}dt)$

I thought that what I should do is just to apply Barrow, and then I'd have:

$\displaystyle\frac{d}{dx}(\displaystyle\int_{2}^{x }\displaystyle\frac{t^{3/2}}{\sqrt[ ]{t^2+17}}dt)=-\displaystyle\frac{x^{3/2}}{\sqrt[ ]{x^2+17}}$

Is this right?

I've tried it on the hard way too, but then:

$t=\sqrt[ ]{17}\sinh u$
$dt=\sqrt[ ]{17}\cosh u du$

$\displaystyle\frac{d}{dx}(\displaystyle\int_{2}^{x }\displaystyle\frac{t^{3/2}}{\sqrt[ ]{t^2+17}dt})=\displaystyle\frac{d}{dx}(\sqrt[3]{17}\displaystyle\int_{2}^{x}\sqrt[3]{\sinh^2 u}du)$

I don't know how to solve the last integration.

Bye there.

2. Originally Posted by Ulysses
The statement says:using the calculus fundamental theorem find:
$\displaystyle\frac{d}{dx}(\displaystyle\int_{2}^{x }\displaystyle\frac{t^{3/2}}{\sqrt[ ]{t^2+17}}dt)$

I thought that what I should do is just to apply Barrow, and then I'd have:

$\displaystyle\frac{d}{dx}(\displaystyle\int_{2}^{x }\displaystyle\frac{t^{3/2}}{\sqrt[ ]{t^2+17}}dt)= \displaystyle\frac{x^{3/2}}{\sqrt[ ]{x^2+17}}$

Is this right?
Yes, this is fine (well, without the negative sign, that is). And this is all you need to do. Forget the hard way, haha

and what's "Barrow"?

3. Isaac Barrow - Wikipedia, the free encyclopedia

Its not needed anyway. I thought of it being x on the "start" of the interval

4. The fundamental theorem of calculus says, in part,
$\frac{d}{dx}\int_a^x f(t)dt= f(x)$

As jhevon said, the negative sign should not be there.