Science:Math Exam Resources/Courses/MATH101 C/April 2024/Question 10/Solution 1

The integrand blows up near $x=1$, so this is an improper integral:

$${\begin{aligned}\int _{0}^{1}{\frac {1}{(1-x)^{3/2}}}\;\mathrm {d} x&=\lim _{t\to 1^{-}}\int _{0}^{t}(1-x)^{-3/2}\;\mathrm {d} x.\end{aligned}}$$

To find the indefinite integral, use the u-substitution $u=1-x$, $\mathrm {d} u=-\mathrm {d} x$:

$${\begin{aligned}\int {\frac {1}{(1-x)^{3/2}}}\;\mathrm {d} x&=-\int u^{-3/2}\mathrm {d} u=-(-2)u^{-1/2}+c=2(1-x)^{-1/2}+c={\frac {2}{\sqrt {1-x}}}+c.\end{aligned}}$$

Thus,

$${\begin{aligned}\int _{0}^{1}{\frac {1}{(1-x)^{3/2}}}\;\mathrm {d} x&=\lim _{t\to 1^{-}}\left.\left[{\frac {2}{\sqrt {1-x}}}\right]\right|_{x=0}^{x=t}=\lim _{t\to 1^{-}}\left({\frac {2}{\sqrt {1-t}}}-2\right)=+\infty .\end{aligned}}$$

The integral diverges.