Determine whether the improper integral converges or diverges.
-$$\int _ { - 9 } ^ { 10 } \frac { d x } { ( x + 1 ) ^ { 1 / 3 } }$$
A) Converges
B) Diverges

Question

Quizplus · Accepted Answer

The integral is improper at $x=-1$, but we can split up the integral into two parts:
\begin{align*}
\int _ { - 9 } ^ { 10 } \frac { d x } { ( x + 1 ) ^ { 1 / 3 } } &= \int _ { - 9 } ^ { -1-\epsilon} \frac { d x } { ( x + 1 ) ^ { 1 / 3 } } + \int _ { -1 + \epsilon} ^ {10} \frac { d x } { ( x + 1 ) ^ { 1 / 3 } }
\end{align*}
where $\epsilon > 0$. We can evaluate each integral separately using the substitution $u = x+1$:
\begin{align*}
\int _ { - 9 } ^ { -1-\epsilon} \frac { d x } { ( x + 1 ) ^ { 1 / 3 } } &= \int _ { -8 } ^ { 0 } \frac {du}{u^{1/3}} = \lim_{t 	o 0^+} 3\left(t^{2/3} - (-8)^{2/3}ight) = 24 \
\int _ { -1 + \epsilon} ^ {10} \frac { d x } { ( x + 1 ) ^ { 1 / 3 } } &= \int _ { 0 } ^ { 11 } \frac {du}{u^{1/3}} = \lim_{t 	o 0^+} 3\left(11^{2/3}-t^{2/3}ight) = 33
\end{align*}
Since both integrals converge, the original integral converges by the comparison test.

Determine Whether the Improper Integral Converges or Diverges ∫−910dx(x+1)1/3\int _ { - 9 } ^ { 10 } \frac { d x } { ( x + 1 ) ^ { 1 / 3 } }∫−910​(x+1)1/3dx​

Determine Whether the Improper Integral Converges or Diverges $\int _ { - 9 } ^ { 10 } \frac { d x } { ( x + 1 ) ^ { 1 / 3 } }$