Question

Integral from 0 to 1 of dx/x^3, does it converge or diverge?

Answer 1

your problem is at 0. find the "ant-derivative" and then see if you can take the limit as t approaches 0

Answer 2

is should be pretty clear that you cannot.

Answer 3

more precisely
\[\int_0^1\frac{dx}{x^3}=\lim_{t\rightarrow 0^+}\int_t^1\frac{dx}{x^3}\]

Answer 4

the anti derivative is
\[-\frac{1}{2x^2}\] so you have
\[\lim_{t\rightarrow 0^+}\frac{1}{2}+\frac{1}{2t^2}\] and this limit does not exist

Answer 5

typo there, first part should be - 1/2, but it makes no difference

Answer 6

Oh okay! I'm just a little confused on the -1/2 part, where did that come from if 1/2t^2 is the anti derivative?

Answer 7

that is what i got when i plugged 1 in to the anti derivative

Answer 8

ooooh!! i get it now.. thanks so much!