Question

Determine if the following integral is convergent or divergent.
${\int }_{-\infty }^0\frac{dx}{\sqrt{3-x}}$

• A. Limit is finite and so this integral is convergent.
• B. Limit is infinite and so this integral is divergent.
• C. Limit does not exist
• D. None of the above

Answer 1

Answer: (B)

Limit is infinite and so this integral is divergent.

Given : ${\int }_{-\infty }^0\frac{dx}{\sqrt{3-x}}$

$\stackrel{lim}{c\to -\infty }{\int }_c^0\frac{dx}{\sqrt{3-x}}=\stackrel{lim}{c\to -\infty }{\left[\frac{2\sqrt{3-x}}{(-1)}\right]}_c^0=\stackrel{lim}{c\to -\infty }\frac{2\sqrt{3}}{(-1)}+\frac{2\sqrt{3-c}}{(+1)}=\infty$

Limit is indefinite and so this integral is divergent.