Proof of Definite Integral?

  • Thread starter Calculuser
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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;

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

Thanks..
 
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  • #2


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, [itex]\int^{b}_{a}f(x)dx[/itex] is equal to [itex]F(b)-F(a)[/itex] because [itex]F(b)[/itex] calculates the area from 0 to b, [itex]F(a)[/itex] calculates the are from 0 to a. Subtracting these two values will cut the area [itex]F(a)[/itex] calculated from [itex]F(b)[/itex] 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.
 
  • #3
chiro
Science Advisor
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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;

[itex]\int^{b}_{a}f(x).dx=F(b)-F(a)[/itex] 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).
 

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