Solving derivative and integral using calculus fundamental theorem

**Ulysses** · July 3rd 2010, 11:16 AM

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.

**Jhevon** · July 3rd 2010, 11:53 AM

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"?

**Ulysses** · July 3rd 2010, 12:47 PM

Isaac Barrow - Wikipedia, the free encyclopedia

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

**HallsofIvy** · July 3rd 2010, 03:43 PM

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.

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