How do you test the improper integral #int (2x-1)^3 dx# from #(-oo, oo)# and evaluate if possible?

1 Answer
Mar 28, 2018

Diverges.

Explanation:

First, let's obtain the indefinite integral #int(2x-1)^3dx# so as to make our work easier when evaluating the improper integral:

#u=2x-1, du=2dx, 1/2du=dx#

#1/2intu^3du=1/8u^4=1/8(2x-3)^4#

No need to include a constant of integration, this was only calculated to keep it on the side for when it becomes needed for evaluating the improper integral.

So,

#int_-oo^oo(2x-3)^3dx=int_-oo^0(2x-3)^3dx+int_0^oo(2x-3)^3dx#

This sort of splitting up is necessary when evaluating from #(-oo, oo).# It's often easier to split up at zero (unless zero is not in the domain of the integrand).

#=lim_(t->-oo)int_t^0(2x-3)^3dx+lim_(t->oo)int_0^t(2x-3)dx#

#=lim_(t->-oo)(1/8(2x-3)^4)|_t^0 + lim_(t->oo)(1/8(2x-3)^4)|_0^t#

#=lim_(t->-oo)(1/8(-3)^4-1/8(2t-3)^4)+lim_(t->oo)(1/8(2t-3)^4-1/8(-3)^4)=-81/8-oo+...#

The first limit did not exist, it went to negative infinity. No need to even evaluate the second one, we already know the entire integral must diverge.