Question

The value of the integral ${\int }_{-1}^1\frac{1}{x^2}dx$

• 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 dis continuity lies with in the [-1,1]
So it is an improper integral of $2^{nd}$ kind
$I={\int }_{-1}^1\frac{1}{x^2}dx={\int }_{-1}^0\frac{1}{x^2}dx+{\int }_0^1\frac{1}{x^2}dx=\infty$