Question

${\int }_1^{\infty }\frac{1}{x^2}dx$ does not have a finite value

• A. False
• B. True

Answer 1

The correct option is A False

Let’s calculate its value.
In such cases where we have to deal with infinity as the limits of definite integral, we’ll change the limit which is not finite to a variable and then put the limits.
${\int }_1^{\infty }\frac{1}{x^2}dx={lim}_{a\to \infty }{\int }_1^a\frac{1}{x^2}dx$
= lim $a\to \infty (-\frac{1}{x}){|}_1^a$ Since, $\int \frac{1}{x^2}dx=(-\frac{1}{x})$
$=0-(-1)=1$