Question

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

Answer 1

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

Answer 2

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

Answer 3

yes

Answer 4

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?

Answer 5

ok got it does it converge

Answer 6

i don't know. do the math.

Answer 7

i think it does converge

Answer 8

ok

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

ok

Answer 13

why b-1

Answer 14

because u have to approach 1 from the left, since u r on the interval [0,1].

it is 1-b, NOT b-1