# Definite Integral w/ Second Fundamental Theorem of Calculus

1. Dec 9, 2007

### lLovePhysics

1. The problem statement, all variables and given/known data
Find F'(x) if

$$F(x)=\int_{0}^{x^3}(\sin (t^2))dt$$

3. The attempt at a solution

Here's what I did:

$$F(x)= -\cos (t^2)\biggr]^{x^3}_{0}$$

and I get: $$F(x)= -\cos (x^6) +1$$

$$F'(x)= sin (x^6)(6x^5)$$

However, the book's answer is $$F'(x)= 3x^2 \sin(x^6)$$

How did they come up with this answer if the derivative of x^6 is 6x^5 ??

Thanks.

Last edited: Dec 9, 2007
2. Dec 9, 2007

### arildno

Eeh, wherever did your cosines come from???

3. Dec 9, 2007

### lLovePhysics

Sorry, original post edited. I was looking at the wrong problem.

4. Dec 9, 2007

### arildno

Allright!
The derivative of -cos(t^{2})=2tsin(t^{2}), which does not agree with your original integrand.

5. Dec 9, 2007

### lLovePhysics

Oh ok. How do you solve it then?

I tired the u-substitution method:

$$F(x)=\frac{1}{2}\int^{x^3}_{0}\sin u(u^{-1/2}) du$$

Am I on the right track?

6. Dec 9, 2007

### arildno

Not at all, you can't write an anti-derivative to that integrand in terms of elementary functions.

Use the fundamental theorem of calculus instead.

7. Dec 10, 2007

### HallsofIvy

Staff Emeritus
You don't find the antiderivative- you don't need to. Use the Fundamental Theorem of Calculus:
$$\frac{d}{dx}\int_a^x f(t)dt= f(x)$$
Handle the fact that the upper limit is a function of x rather than x itself with the chain rule:
[tex]\frac{d}{dx}\int_a^{g(x)}f(t)dt= f(x)\frac{dg}{dx}[/itex]