Question

Use limits of definite integrals to calculate each of the improper integrals in Exercises.

${\int }_1^{\infty }\frac{1}{{\left(x-2\right)}^2}dx$

Short answer

The improper integral ${\int }_1^{\infty }\frac{1}{{\left(x-2\right)}^2}dx$diverges.

Step-by-step solution

Step 1. Given information. 

The given integral is the following.

${\int }_1^{\infty }\frac{1}{{\left(x-2\right)}^2}dx$

Step 2. Value of integral. 

The function $f\left(x\right)=\frac{1}{{\left(x-2\right)}^2}$has a vertical asymptote at$x=2.$

solve this improper integral by splitting at $x=2$

role="math" localid="1648878472203" ${\int }_1^{\infty }\frac{1}{{\left(x-2\right)}^2}dx={\int }_1^2\frac{1}{{\left(x-2\right)}^2}dx+{\int }_2^{\infty }\frac{1}{{\left(x-2\right)}^2}dx$

Take role="math" localid="1648878379284" ${\int }_1^2\frac{1}{{\left(x-2\right)}^2}dx$

role="math" localid="1648878372798" $$\begin{gathered} {\int }_0^2\frac{1}{{\left(x-2\right)}^2}dx=\underset{b\to 0}{\lim }{\int }_b^2\frac{1}{{\left(x-2\right)}^2}dx \\ =\underset{b\to 0}{\lim }{\left[\frac{1}{\left(x-2\right)}\right]}_b^2 \\ =\underset{b\to 0}{\lim }\left(1/0-\frac{1}{\left(b-2\right)}\right) \\ =\frac{1}{0}+\frac{1}{2}=\mathrm{undefined} \end{gathered}$$

Integral ${\int }_1^2\frac{1}{{\left(x-2\right)}^2}dx$is undefined and diverges.

So integral ${\int }_1^{\infty }\frac{1}{{\left(x-2\right)}^2}dx$is also diverged.