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

Question

anonymous · Answer

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

amistre64 · Answer

looks to be missing a 2

amistre64 · Answer

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

amistre64 · Answer

$$\frac{-1}{2(2.0-1)^2}-(\frac{-1}{2(2.inf-1)^2})$$

$$\frac{-1}{2(-1)^2}-(\frac{-1}{2(inf)^2})$$

$$\frac{-1}{2}-(\frac{-1}{inf})$$

$$\frac{-1}{2}-(0)$$

across · Answer

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

amistre64 · Answer

4? really?

across · Answer

I performed the substitution$$u=2x-1,$$$$du=2dx.$$

amistre64 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

so yeah, -1/4 is it ;)

anonymous · Answer

Thank You! :)