Integral formula?

[khenemetre]

The formula for the derivative is

 

lim (f(x+h)-f(x))/h

h->0

 

But is there a similar formula for finding indefinite integrals? Help!

[Bignose]

Riemann sums work for definitie integrals

 

[math]\int_{a}^{b} f(x)dx = \lim_{\max \Delta x_{k} \rightarrow 0} \sum_{k=1}^{n} f(x^{*}_{k}) \Delta x_{k} [/math]

 

[math]x^{*}_{k}[/math] is an arbirtrary point in the interval [math]\Delta x_{k} [/math]

[D H]

Riemann sums work for definitie integrals

 

That doesn't answer the original question.

 

The short answer is "no".

 

A smart-alecky answer is the Fundamental Theorem of Calculus,

[math]F(x) = \int f(x) dx \Leftrightarrow f(x) = \frac {dF(x)}{dx}[/math]

 

In other words, you know you have the right answer if the derivative of that answer is the function to be integrated. How to get that right answer, though, is tricky. There is no general rule on how to integrate a function. You need to learn many rules and learn which of those rules is of use on a particular problem.

[Bignose]

I took a slightly different tack on the question -- probably not what the OP was looking for now that I think about it. I look the limit of the discrete approximations as it approachs the continuous case. I think that this means I've been doing too much computational analysis lately!

 

Actually, I may have been as far off as originally thought. Look at the example done for f(x)=x^2 on Mathworld:

 

http://mathworld.wolfram.com/RiemannIntegral.html

 

It is a lot of work, and there are far easier ways to get the integral, but the Reimann sum will provide the right answer. I am sure there are restrictions as to when it will work, but in most cases where the integral is needed, the Reimann integral does its job. Now, as to indefinite integrals, then you need the Fundamental Theorem, but again I was thinking in terms of the limit as discrete jumps go to zero.

[sinisterwolf]

Actually instead of using the dx=f(x+h)-f(x) as h---->infiniti you can also use this....if ax^n then dx=(an)x^(n+1) so then if you want to do an intergral then instead of adding one to the exponent you subtract one and then multiply that by the recipicol of the exponent....if you understand what i just said then try it it works I swear....

[Ragib]

Umm well first, its df/dx, and its as h -> 0. I have no idea what your method is trying to show, but that only works for polynomials anyway. The OP wants a general one for all functions.