Q. 2TF - Calculus

Question

Improper Integrals: Determine whether the following improper integrals converge or diverge.

${\int }_1^{\infty }\frac{1}{x^2}dx.$

Step-by-Step Solution

Answer

The series is a convergent series.

Step 1. Given Information.

$Given an integral: {\int }_1^{\infty }\frac{1}{x^2}dx.$

Step 2. Solving the integral.

$$\begin{gathered} {\int }_1^{\infty }\frac{1}{x^2}dx \\ = {\left[\frac{-1}{x}\right]}_1^{\infty } \\ = \left[\frac{-1}{\infty }-\left(\frac{-1}{1}\right)\right] \\ = \left[0+1\right] = 1. \\ Therefore, the series is a convergent series. \end{gathered}$$