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## Homework Statement

Prove that

[tex]f(x)g(x)-f(0)g(0)=\int^{x}_{0}f(t)g'(t)dt+f'(t)g(t)dt[/tex]

## Homework Equations

Basically, the question also tells you that you should somehow use the fundamental theorem of calculus.

## 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:

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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

[tex]let G(x)=f(x)g(x)[/tex]

[tex]then G'(x)=f'(x)g(x)+f(x)g'(x)[/tex]

[tex]\therefore f(x)g(x)-f(0)g(0) = G(x) - G(0)[/tex]

[tex]\int^x_0 f'(t)g(t)dt+f(t)g'(t)dt[/tex]

[tex]=\int^x_0 G'(t)dt[/tex]

fundamental theorem of calculus states that:

[tex]\int^x_a G'(t)dt = G(x)|^x_a[/tex]

[tex]\therefore \int^x_0 G'(t)dt = G(x)|^x_0=G(x)-G(0) = f(x)g(x)-f(0)g(0)[/tex]

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

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