Is there a general way to solve integrals?

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Homework Statement



This is just something I've been wondering, but since derivatives have the formula:

dy/dx = lim h-> 0 of (f(x+h) - f(x)/h)

And that formula can prove a lot of derivatives.

Does a similar formula exist that can prove integrals?

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The Attempt at a Solution

 
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Yes. Check out riemann sums.

You take the limits of areas of rectangles under the curve as the number of rectangles approaches infinity.
 
CalculusHelp1 said:
Yes. Check out riemann sums.

You take the limits of areas of rectangles under the curve as the number of rectangles approaches infinity.

This is actually the original way that integrals were computed before the discovery of the Fundamental Theorem of Calculus.
 
Would it be possible to prove most integrals without the FTC using Riemann Sums?
 
Voivode said:
Would it be possible to prove most integrals without the FTC using Riemann Sums?

Well, first you'd have to define what you mean by "most integrals". After all, there are an infinite amount of integrals with a solution in the elementary functions. I am relatively certain that all polynomial integrals can be solved in this fashion.
 
By most integrals, I was mainly thinking of the ILATE functions. Could those be solved with riemann sums?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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