Proving Integral of a Product

In summary, the task is to prove that the given equation holds true, using the fundamental theorem of calculus. The solution involves defining a function G(x) as f(x)g(x) and using the derivative of G(x) to simplify the equation. Ultimately, the proof is based on the fact that the integral of a function's derivative is equal to the difference of the function at the upper and lower limits of integration.
  • #1
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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:rolleyes:

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

[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.
 
Last edited:
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  • #2
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.
 

1. How do you prove the integral of a product?

To prove the integral of a product, you need to use the integration by parts formula: ∫u(x)v'(x)dx = u(x)v(x) - ∫v(x)u'(x)dx. This formula allows you to break down the product into two separate functions and integrate them separately.

2. What is the purpose of proving the integral of a product?

The purpose of proving the integral of a product is to find a way to calculate the integral of a product of two functions. This is useful in many areas of science, such as physics and engineering, where it is common to work with products of functions.

3. Can the integral of a product be proven using other methods?

Yes, the integral of a product can also be proven using substitution, trigonometric identities, or other integration techniques depending on the specific functions involved. However, the integration by parts method is the most commonly used and effective approach.

4. Are there any special cases when proving the integral of a product?

Yes, there are some special cases to consider when proving the integral of a product. For example, if one of the functions is a constant, the integration by parts formula simplifies to ∫u(x)dx = u(x)x + C. Another special case is when one of the functions is an exponential, in which case the formula becomes ∫e^u(x)dx = e^u(x) + C.

5. How do you know when to use integration by parts to prove the integral of a product?

You should use integration by parts when the integral contains a product of two functions that can't be easily integrated using other methods. Usually, this involves one function that is getting more complicated when differentiated and another function that gets simpler when integrated.

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