# Math Help - Improper Integral with one discontinuous integrand

1. ## Improper Integral with one discontinuous integrand

$\int_0^2 \ \frac{1}{\sqrt[3]{x-1}}\, dx.$

I know that it is undefined at x=1 because the denominator can't be 0, so I did this:
$\lim_{x \rightarrow 1-} \int_0^c \ \frac{1}{\sqrt[3]{x-1}}\,$ + $\lim_{x \rightarrow 1+} \int_c^2 \ \frac{1}{\sqrt[3]{x-1}}\,$

Then I integrated it...
$\lim_{x \rightarrow 1-} \frac{3}{2}\ (x-1)^\frac{2}{3}$ $|_0^C$+ $\lim_{x \rightarrow 1+} \frac{3}{2}\ (x-1)^\frac{2}{3}$ $|_C^2$

Plugging everything in-

$\lim_{x \rightarrow 1-} \frac{3}{2}\ (c-1)^\frac{2}{3}$ - $\frac{3}{2}\ (0-1)^\frac{2}{3}$ + $\lim_{x \rightarrow 1+} \frac{3}{2}\ (2-1)^\frac{2}{3}$ - $\frac{3}{2}\ (c-1)^\frac{2}{3}$

Then I started to find the limit but somehow I got the wrong answer. I got 1.5-1.5+1.5-1.5 for some reason, because I got that the one sided limits go to 1.5. The answer is supposed to be 3.

2. Originally Posted by juicysharpie
$\int_0^2 \ \frac{1}{\sqrt[3]{x-1}}\, dx.$

I know that it is undefined at x=1 because the denominator can't be 0, so I did this:
$\lim_{x \rightarrow 1-} \int_0^c \ \frac{1}{\sqrt[3]{x-1}}\,$ + $\lim_{x \rightarrow 1+} \int_c^2 \ \frac{1}{\sqrt[3]{x-1}}\,$

Then I integrated it...
$\lim_{x \rightarrow 1-} \frac{3}{2}\ (x-1)^\frac{2}{3}$ $|_0^C$+ $\lim_{x \rightarrow 1+} \frac{3}{2}\ (x-1)^\frac{2}{3}$ $|_C^2$

Plugging everything in-

$\lim_{x \rightarrow 1-} \frac{3}{2}\ (c-1)^\frac{2}{3}$ - $\frac{3}{2}\ (0-1)^\frac{2}{3}$ + $\lim_{x \rightarrow 1+} \frac{3}{2}\ (2-1)^\frac{2}{3}$ - $\frac{3}{2}\ (c-1)^\frac{2}{3}$

Then I started to find the limit but somehow I got the wrong answer. I got 1.5-1.5+1.5-1.5 for some reason, because I got that the one sided limits go to 1.5. The answer is supposed to be 3.

$\lim_{c \rightarrow 1^{-}} \frac{3}{2}\ (c-1)^\frac{2}{3} = 0$. Similarly for the second one.