$\text { Determine whether the improper integral } \int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt [ 3 ] { 1 - x } } d x \text { diverges or converges. }$ Evaluate the integral if it converges.
A) 3
B) $\frac { 2 } { 3 }$
C) $\frac { 3 } { 2 }$
D) 2
E) diverges

Question

$\text { Determine whether the improper integral } \int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt [ 3 ] { 1 - x } } d x \text { diverges or converges. }$ Evaluate the integral if it converges. 
A) 3
B) $\frac { 2 } { 3 }$
C) $\frac { 3 } { 2 }$
D) 2
E) diverges

Quizplus · Accepted Answer

The integral converges. By substituting $ u = 1 - x $, the integral becomes $\int_{1}^{0} \frac{-1}{u^{1/3}} du$, which simplifies to $\int_{0}^{1} u^{-1/3} du$. Evaluating this gives $\left[ \frac{u^{2/3}}{2/3} \right]_{0}^{1} = \frac{3}{2}$.