## Calculus: Early Transcendentals 8th Edition

$$\int_{0}^{9} \frac{1}{\sqrt[3]{x-1}} d x$$ We note first that the given integral is improper because $f(x)= \frac{1}{\sqrt[3]{x-1}}$ has the vertical asymptote $x=1$. The infinite discontinuity occurs at the middle of the interval $[0,9]$ at $x = 1$. The given improper integral is convergent and its value is $$\int_{0}^{9} \frac{1}{\sqrt[3]{x-1}} d x = \frac{9}{2}.$$
$$\int_{0}^{9} \frac{1}{\sqrt[3]{x-1}} d x$$ Observe that the given integral is improper because $f(x)= \frac{1}{\sqrt[3]{x-1}}$ has the vertical asymptote $x=1$. Since the infinite discontinuity occurs at the middle of the interval $[0,9]$ at $x = 1$, we must use part (c) of Definition 3 with $c=1$: $$\int_{0}^{9} \frac{1}{\sqrt[3]{x-1}} d x=\int_{0}^{1}(x-1)^{-1 / 3} d x+\int_{1}^{9}(x-1)^{-1 / 3} d x \quad (1)$$ where $$\begin{split} \int_{0}^{1}(x-1)^{-1 / 3} d x & = \lim _{t \rightarrow 1^{-}} \int_{0}^{t}(x-1)^{-1 / 3} d x \\ & =\lim _{t \rightarrow 1^{-}}\left[\frac{3}{2}(x-1)^{2 / 3}\right]_{0}^{t} \\ & =\lim _{t \rightarrow 1^{-}}\left[\frac{3}{2}(t-1)^{2 / 3}-\frac{3}{2}\right] \\ & =-\frac{3}{2} \end{split} \quad \quad\quad (2)$$ and $$\begin{split} \int_{1}^{9}(x-1)^{-1 / 3} d x & = \lim _{t \rightarrow 1^{+}} \int_{t}^{9}(x-1)^{-1 / 3} d x \\ & =\lim _{t \rightarrow 1^{+}}\left[\frac{3}{2}(x-1)^{2 / 3}\right]_{t}^{9} \\ & =\lim _{t \rightarrow 1^{+}}\left[6-\frac{3}{2}(t-1)^{2 / 3}\right] \\ & =6 \end{split} \quad \quad\quad (3)$$ Substituting the integrals (2) and (3) into expression (1) , we obtain $$\int_{0}^{9} \frac{1}{\sqrt[3]{x-1}} d x=-\frac{3}{2} +6 =\frac{9}{2},$$ It follows that the integral $$\int_{0}^{9} \frac{1}{\sqrt[3]{x-1}} d x$$ is convergent.