# Proving Integral of a Product

1. Nov 30, 2006

### bluevires

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

Prove that
$$f(x)g(x)-f(0)g(0)=\int^{x}_{0}f(t)g'(t)dt+f'(t)g(t)dt$$
2. Relevant equations
Basically, the question also tells you that you should somehow use the fundamental theorem of calculus.

3. The attempt at a solution
This is pretty much asking me to prove that the derivative of f(x)g(x) is just f'(x)g(x) + f(x)g'(x)...and then I could use the fundamental theorem of calculus to say that the integral of the derivative is just the function itself.

It seems pretty self-explanatory to me, so I havn't got a clue on where to start on this proof, I would appreciate a tip or two to help me get started

Update:
--------------------
here's what I got so far. I don't know if this is right or wrong, so I would appreciate some comment on it

$$let G(x)=f(x)g(x)$$
$$then G'(x)=f'(x)g(x)+f(x)g'(x)$$

$$\therefore f(x)g(x)-f(0)g(0) = G(x) - G(0)$$
$$\int^x_0 f'(t)g(t)dt+f(t)g'(t)dt$$
$$=\int^x_0 G'(t)dt$$
fundamental theorem of calculus states that:
$$\int^x_a G'(t)dt = G(x)|^x_a$$
$$\therefore \int^x_0 G'(t)dt = G(x)|^x_0=G(x)-G(0) = f(x)g(x)-f(0)g(0)$$

sorry if the equations are not aligned properly, I'm a newbie at LaTex, thx all.

Last edited: Nov 30, 2006
2. Nov 30, 2006

### AlephZero

Well, I think that's pretty much all you need to do.

Just as a detail on how to write it out, I would leave out the 3rd line "therefore f(x)g(x) ....." because you the same thing again in the last line, and it comes straight from your definition of the function G so it's pretty obvious.

When you say "therefore" on line 3 it might make a reader think that line 3 follows from line 2, which is misleading.

The key point (as you said) is to notice that f(x)g'(x) + f'(x)g(x) is the derivative of f(x)g(x). The rest is just notation, really.