Can Squaring Integrals Simplify Calculus Problems?

[Ibraheem]

Hello,

When I recently was studying for my calculus I's rules of definite integrals, I was wondering if squaring a definite integral would be the same as integrating twice like in the following:( Definite integral of f(x) from a to b)^2 = definite integral ,from a to b, of the definite integral of f(x) from a to b)

 

[PeroK]

Hello,

When I recently was studying for my calculus I's rules of definite integrals, I was wondering if squaring a definite integral would be the same as integrating twice like in the following:( Definite integral of f(x) from a to b)^2 = definite integral ,from a to b, of the definite integral of f(x) from a to b)

Did you think to try this with an example function f(x)?

 

[mathman]

Simplest example f(x)=x.

 

[Ibraheem]

Yes, It is clearer now.

( Definite integral of f(x) from a to b)^2 does not equal definite integral ,from a to b, of the definite integral of f(x) from a to b)

 

[WWGD]

It may be interesting to figure out if it is ever actually true. But it is a hard problem, considering all possible functions and all possible limits of integration. Seems maybe an issue of calculus of variations?

 

[davidmoore63@y]

How does this make sense? The RHS is "definite integral ,from a to b, of the definite integral of f(x) from a to b)"
which equals definite integral, from a to b, of a constant.

 

[WWGD]

A constant is a continuous function, which can be Riemann-integrated: $\int_a^b C dx=C(b-a)$

 

[davidmoore63@y]

ok then so letting Integral[a,b] f(x)dx =C, the problem is equivalent to C^2= C(b-a) and hence C=0 or b-a

 

[WWGD]

Yes, seems right.

 

[Michael David]

Hmm I think i saw something like what you're saying once.. a way of solving the poisson integral by squaring it... It was something like I^2=Integral [minus, plus infinity] e^(-x^2)dx times same thing but with y instead of x... then you get integral [minus,plux] of integral [minus,plus] of e^-(x^2+y^2)dxdy which is an integral over the whole xy plane... and so you can change to polar coordinates and you easily get the result for that integral which is pi... so there you have it.. I squared is pi, therefor I is the square root of pi..