How Can the Mean Value Theorem Prove the Derivative of an Indefinite Integral?

[courtrigrad]

Hello all

Using the Mean Value Theorem, prove that the derivative of the indefinite integral \int f(x) \ dx is f(x)

So do I just use the fact that \int^b_a f(x) \ dx = f(\xi)(b-a)?

Thanks

 

[courtrigrad]

is this right?

 

[HallsofIvy]

You CAN'T use the mean value theorem to prove the Fundamental Theorem of Calculus.

 

[courtrigrad]

not prove but maybe show

 

[vincentchan]

hey... do you forget our rule here... don't give out the answer... delete the link and give him some hints lead to the answer

 

[learningphysics]

hey... do you forget our rule here... don't give out the answer... delete the link and give him some hints lead to the answer

Yes,you're right sorry about that.

 

[learningphysics]

I believe your question should to prove that the derivative of this function:

F(x)=\int_a^{x} f(t)dt

is f(x). Am I right? The above is a definite integral. The derivative of the indefinite integral is just f(x) by definition. Indefinite integral means anti-derivative.

What is F'(x) from first principles i.e: using the definition of derivative?

 

[courtrigrad]

F'(x) = \frac{F(x+\Delta x) - F(x)}{\Delta x}

forgot to put limit

 

[learningphysics]

F'(x) = \frac{f(x+\Delta x) - f(x)}{\Delta x}

forgot to put limit

Careful... we're looking for F'(x) not f'(x).

 

[courtrigrad]

is this right?

 

[learningphysics]

is this right?

Yes. So now use the definition of F(x) in post #7, to plug in the approriate F(x) and F(x+deltax) into your derivative equation...