Constant in integration

[jpanknin]

I have the following integral:



My question is that in the second line, r^2 is a constant. Why wasn't it taken out and moved before the integral sign along with 2pi? I understand that r^2-x^2 = y^2, but this seems to violate the rule of moving constants in front of the integral sign.

 

[Dr.Peterson]

I have the following integral:

View attachment 31115

My question is that in the second line, r^2 is a constant. Why wasn't it taken out and moved before the integral sign along with 2pi? I understand that r^2-x^2 = y^2, but this seems to violate the rule of moving constants in front of the integral sign.

There is no rule that you must move constants in front of the integral sign.

It is a appropriate to do so for constant factors of the entire integrand; but [imath]r^2[/imath] is not a factor; it is a term. As such, it still needs to be integrated. (You could take it outside when you integrate the first term, leaving an integrand of 1 for that term; but that would require writing more.)

 

[nasi112]

If the integral is like this, you can move the constant out,

\(\displaystyle 2\pi\int_0^r (r^2 - r^2x^2) \ dx = 2\pi r^2\int_0^r (1 - x^2) \ dx\)

In your case, you can also move it out, but you have to do it like this,

\(\displaystyle 2\pi\int_0^r (r^2 - x^2) \ dx = 2\pi r^2\int_0^r \left(1 - \frac{x^2}{r^2}\right) \ dx\)

which is not necessary as it did not help to simplify anything.

 

[Steven G]

\(\displaystyle 2\pi\int_0^r (r^2 - x^2) \ dx = 2\pi[r^2\int_0^r 1 \ dx - \int_0^r\ x^2 \ dx] \)