Answer
$$
\int_{-1}^{0} \frac{e^{1 / x}}{x^{3}} d x
$$
Observe that the given integral is improper because $ f(x)= \frac{e^{1 / x}}{x^{3}} $ has the vertical asymptote $x =0 $. The infinite discontinuity occurs at the right endpoint of $ [-1, 0] $.
Thus the given improper integral
$$
\int_{-1}^{0} \frac{e^{1 / x}}{x^{3}} d x=-\frac{2}{e}
$$
is convergent
Work Step by Step
$$
\int_{-1}^{0} \frac{e^{1 / x}}{x^{3}} d x
$$
Observe that the given integral is improper because $ f(x)= \frac{e^{1 / x}}{x^{3}} $ has the vertical asymptote $\theta =0 $. Since the infinite discontinuity occurs at the right endpoint of $ [-1, 0] $ , we must use part (a) of Definition 3 :
$$
\begin{split}
\int_{-1}^{0} \frac{e^{1 / x}}{x^{3}} d x & = \lim _{t \rightarrow 0 ^{-}} \int_{-1}^{t} \frac{e^{1 / x}}{x^{3}} d x \\
& =\lim _{t \rightarrow 0^{-}} \int_{-1}^{t} \frac{1}{x} e^{1 / x} \cdot \frac{1}{x^{2}} d x \\
& =\lim _{t \rightarrow 0^{-}} \int_{-1}^{1 / t} u e^{u}(-d u) \quad\left[\begin{array}{c}{u=1 / x} \\ {d u=-d x / x^{2}}\end{array}\right] \\
& =-\lim _{t \rightarrow 0^{-}} \int_{-1}^{1 / t} u e^{u}(d u),
\quad\quad \text{integration by parts } \\
& =-\lim _{t \rightarrow 0^{+}}\left[(u-1) e^{u}\right]_{1}^{1 / t} \\
& = \lim _{t \rightarrow 0^{-}}\left[-2 e^{-1}-\left(\frac{1}{t}-1\right) e^{1 / t}\right] \\
& =-\frac{2}{e}-\lim _{s \rightarrow-\infty}(s-1) e^{s} \quad[s=1 / t] \\
& = -\frac{2}{e}-\lim _{s \rightarrow-\infty} \frac{s-1}{e^{-s}} \quad\quad \text{ use l’Hospital’s Rule } \\
& = -\frac{2}{e}-\lim _{s \rightarrow-\infty} \frac{1}{-e^{-s}} \\
& = -\frac{2}{e}-0=-\frac{2}{e}
\end{split}
$$
Thus the given improper integral
$$
\int_{-1}^{0} \frac{e^{1 / x}}{x^{3}} d x=-\frac{2}{e}
$$
is convergent
