How to prove the fundamental theorem of calculus

[Calculuser]

Proof of Definite Integral??

Riemann Sum helps us to find the are under a curve.But I couldn't understand connection between Riemann Sum and Definite Integral.I mean;

$\int^{b}_{a}f(x).dx=F(b)-F(a)$ How can I prove F(b)-F(a) in definite integral??

Thanks..

 

[Shootertrex]

Well, the connection between the two is that the definite integral is essentially using an infinite amount of Riemann sums to calculate the area under a curve. Integration allows for the area to be calculated without having to add up the area of each Riemann sum manually.

Now, $\int^{b}_{a}f(x)dx$ is equal to $F(b)-F(a)$ because $F(b)$ calculates the area from 0 to b, $F(a)$ calculates the are from 0 to a. Subtracting these two values will cut the area $F(a)$ calculated from $F(b)$ leaving just the area from a to b.

I hope I was able to provide the information that you needed for you to finish your proof.

 

[chiro]

Riemann Sum helps us to find the are under a curve.But I couldn't understand connection between Riemann Sum and Definite Integral.I mean;

$\int^{b}_{a}f(x).dx=F(b)-F(a)$ How can I prove F(b)-F(a) in definite integral??

Thanks..

Hey Calculuser and welcome to the forums.

You actually can't do this for any f(x): it needs to what is called "Riemann-Integrable" for this to happen.

There are strict conditions that can be representing through formal mathematical statements but the basic idea is that your anti-derivative must be analytically differentiable and continuous everywhere across the domain (i.e. from a to b).

If this is not the case, then you have to either separate your intervals into ones where this happens or use a different integration technique (like for when you are modelling things like Brownian motion).