Problem With fundamental theorum of Calculus

[Titans86]

Homework Statement

The question: F'($$\pi$$/2) if F(X)= $$\int^{cosx}_{0} e^{t^{2}}$$

The Attempt at a Solution

I thought I thought F'(X) = f(t) = e^{t^{2}} replacing t with cos^{2}x

But my book writes:

F'(x) = $$(-sinx)e^{cos^{2}x}$$

 

[sutupidmath]

the deal is that you need to apply chain rule here, since your upper limit bound of integration is not an independent variable, but rather a function as well.

 

[Avodyne]

I assume your question is, "Find the value of $F'(\pi/2)$, given

$$F(x)=\int_0^{\cos x}e^{t^2}\,dt\;.$$

The fundamental theorem of calculus states that if

$$F(x)=\int_a^x f(t)dt\;,$$

then

$$F'(x)=f(x).$$

Can you see how this differs from your problem?

 

[Titans86]

hmm... I'm sorry but I still don't see it...

The book then substitutes pi/2 into the F'(x) that I wrote above and ends it there...

the final answer is -1...

 

[sutupidmath]

hmm... I'm sorry but I still don't see it...

The book then substitutes pi/2 into the F'(x) that I wrote above and ends it there...

the final answer is -1...

What does the chain rule say?

if you have$$[F(g(x))]'=F'(g(x))*g'(x)=e^{cos^2x}(cosx)'=?$$

In your problem you have g(x)=cosx and

$$F(g(x))=\int_0^{\ g(x)}e^{t^2}\,dt=\int_0^{\cos x}e^{t^2}\,dt$$

Now just look above and you are done.

 

[Titans86]

Ah, I see what's going on...

I was differentiating $$e^{cos^{2}x}$$ which was giving me something else...