Answer
The given improper integral is:
$$
\int_{0}^{1} \frac{x-1}{\sqrt{x}} dx = -\frac{4}{3}
$$
Work Step by Step
$$
\int_{0}^{1} \frac{x-1}{\sqrt{x}} dx
$$
Observe that the given integral is improper because $ f(x)=\frac{x-1}{\sqrt{x}} $ has the vertical asymptote $x =0 $. Since the infinite discontinuity occurs at the left endpoint of $ [0, 1] $ , we must use part (b) of Definition 3 :
$$
\begin{aligned} \int_{0}^{1} \frac{x-1}{\sqrt{x}} d x &=
\lim _{t \rightarrow 0^{+}} \int_{t}^{1}\left(\frac{x}{\sqrt{x}}-\frac{1}{\sqrt{x}}\right) d x
=\lim _{t \rightarrow 0^{+}} \int_{t}^{1}\left(x^{1 / 2}-x^{-1 / 2}\right) d x=\lim _{t \rightarrow 0^{+}}\left[\frac{2}{3} x^{3 / 2}-2 x^{1 / 2}\right]_{t}^{1}
\\ &=\lim _{t \rightarrow 0^{+}}\left[\left(\frac{2}{3}-2\right)-\left(\frac{2}{3} t^{3 / 2}-2 t^{1 / 2}\right)\right]=-\frac{4}{3}-0=-\frac{4}{3} \end{aligned}
$$
\begin{aligned} \int_{0}^{1} \frac{x-1}{\sqrt{x}} d x &=\lim _{t \rightarrow 0^{+}} \int_{t}^{1}\left(\frac{x}{\sqrt{x}}-\frac{1}{\sqrt{x}}\right) d x \\
&=\lim _{t \rightarrow 0^{+}} \int_{t}^{1}\left(x^{1 / 2}-x^{-1 / 2}\right) d x \\
&= \lim _{t \rightarrow 0^{+}}\left[\frac{2}{3} x^{3 / 2}-2 x^{1 / 2}\right]_{t}^{1} \\
&= \lim _{t \rightarrow 0^{+}}\left[\left(\frac{2}{3}-2\right)-\left(\frac{2}{3} t^{3 / 2}-2 t^{1 / 2}\right)\right]\\
&= -\frac{4}{3}-0=-\frac{4}{3}
\end{aligned}
Thus, the given improper integral is:
$$
\int_{0}^{1} \frac{x-1}{\sqrt{x}} dx = -\frac{4}{3}
$$
