Question

The value of integral $\underset{-1}{\overset{1}{\int }}\frac{1}{x^2}dx$ is

• A. 2
• B. does not exist
• C. -2
• D. $\infty$

Answer 1

The correct option is D $\infty$

$\because$ x= 0 is a point of discontinuity lies within the range [-1, 1].
So, it is improper integral of $2^{nd}$ kind.
$I=\underset{-1}{\overset{1}{\int }}\frac{1}{x^2}dx=\underset{-1}{\overset{0}{\int }}\frac{1}{x^2}dx+\underset{0}{\overset{1}{\int }}\frac{1}{x^2}dx=\infty$