# Problem With fundamental theorum of Calculus

1. Dec 17, 2008

### Titans86

1. The problem statement, all variables and given/known data

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

3. 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) = $$(-sinx)e^{cos^{2}x}$$

2. Dec 17, 2008

### 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.

3. Dec 17, 2008

### 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?

4. Dec 17, 2008

### 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...

5. Dec 17, 2008

### sutupidmath

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.

Last edited: Dec 17, 2008
6. Dec 17, 2008

### Titans86

Ah, I see whats going on...

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