Question

Evaluate the integral from -infinty to 0 (1/(2x-1)^3) and ttell whether it diverges or converges.

Answer 1

\[\int\limits_{-\infty}^{0} 1/(2x-1)^{3}\]

Answer 2

looks to be missing a 2

Answer 3

\[\frac{-1}{2(2x-1)^2}\]
is the integral part i believe

Answer 4

\[\frac{-1}{2(2.0-1)^2}-(\frac{-1}{2(2.inf-1)^2})\]

\[\frac{-1}{2(-1)^2}-(\frac{-1}{2(inf)^2})\]

\[\frac{-1}{2}-(\frac{-1}{inf})\]

\[\frac{-1}{2}-(0)\]

Answer 5

\[\int_{-\infty}^{0}\frac{1}{(2x-1)^3}dx=\left [ -\frac{1}{4(1-2x)^2} \right ]_{-\infty}^{0}\]\[\implies -\frac{1}{4}+0=-\frac{1}{4}.\]Seems to converge.

Answer 6

4? really?

Answer 7

I performed the substitution\[u=2x-1,\]\[du=2dx.\]

Answer 8

\[dy/dx\ \frac{-1}{2}(2x-1)^{-2} = \frac{2}{2}(2x-1)^{-3}\]

Answer 9

opps, forgot the 2 that pops out of the inners :)

Answer 10

so yeah, -1/4 is it ;)

Answer 11

Thank You! :)