Derivative and integral using calculus foundamental theorem

[Telemachus]

Homework Statement

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

The Attempt at a Solution

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.

 

[LCKurtz]

Well, you have at least one typo; do you have two of them? The dt should not be in the denominator. And is the d/dt at the beginning typed correctly or is it supposed to be d/dx?

 

[Telemachus]

I've corrected the typos. Thanks LCKurtz. Its d/dt.

 

[l'Hôpital]

It doesn't make any sense. Are you sure you don't mean d/dx? Otherwise, the answer is zero, assuming x is not a function t.
$$\displaystyle\frac{d}{dt}(\displaystyle\int_{2}^{x }\displaystyle\frac{t^{3/2}}{\sqrt[ ]{t^2+17}}dt) = \frac{d}{dt} (F(x) - F(2) ) = 0$$

 

[Telemachus]

Sorry, you're right. I'll edit right now.

It doesn't make any sense. Are you sure you don't mean d/dx? Otherwise, the answer is zero, assuming x is not a function t.
$$\displaystyle\frac{d}{dt}(\displaystyle\int_{2}^{x }\displaystyle\frac{t^{3/2}}{\sqrt[ ]{t^2+17}}dt) = \frac{d}{dt} (F(x) - F(2) ) = 0$$

How can I find F(x)?

 

[vela]

If you use the fundamental theorem of calculus, you don't need to find F(x). That's the point of the problem.

 

[Telemachus]

Thanks.

 

[Mark44]

Homework Statement

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

The Attempt at a Solution

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?

You might have this completely figured out by now, but just in case you haven't, that minus sign shouldn't be there on the right side of the equation.

 

[Telemachus]

Yep, at the beginning I thought of it as $$\displaystyle\frac{d}{dx}(\displaystyle\int_{x}^{2}\displaystyle\frac{t^{3/2}}{\sqrt[ ]{t^2+17}}dt)=$$. Now I've realized that it was wrong :P

Thanks to all of you people.