determine whether convergent of divergent and evaluate integral of (x-1)^(-2/3)dx from 0 to 9

Question

anonymous · Answer

$$\int\limits_{0}^{9}1/ \sqrt[3]{x^2-1}dx$$
thT IS THE eq

helder_edwin · Answer

actually it is
$$\large \int_0^9\frac{1}{(x-1)^{2/3}}\,dx $$

anonymous · Answer

yes

helder_edwin · Answer

the function has a discontinuity at x=1. split the integral into two integral 0 to 1 and 1 to 9. and use limits.

got it?

anonymous · Answer

ok got it does it converge

helder_edwin · Answer

i don't know. do the math.

helder_edwin · Answer

i think it does converge

anonymous · Answer

ok

helder_edwin · Answer

u should do
$$\large \int_0^9(x-1)^{-2/3}\,dx=\int_0^1(x-1)^{-2/3}\,dx+
\int_1^9(x-1)^{-2/3}\,dx $$

helder_edwin · Answer

$$\large \int_0^1(x-1)^{-2/3}\,dx=\lim_{b\to0+}\int_0^{1-b}(x-1)^{-2/3}\,dx$$

helder_edwin · Answer

do the same with the other with 1+b instead of 1-b

anonymous · Answer

ok

anonymous · Answer

why b-1

helder_edwin · Answer

because u have to approach 1 from the left, since u r on the interval [0,1].

it is 1-b, NOT b-1