Question

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

${\int }_1^{\infty }{x}^{-0.99}dx$

Short answer

The improper integral on $[1,\infty )$diverges.

${\int }_1^{\infty }{x}^{-0.99}dx=\infty$

Step-by-step solution

Step 1. Given information.

The given improper integral is following.

${\int }_1^{\infty }{x}^{-0.99}dx$

Step 2. Value of integral.

Write the integral in the proper form of ${\int }_1^{\infty }\frac{1}{x^p}dx$

$$\begin{gathered} {\int }_1^{\infty }\frac{1}{x^p}dx={\int }_1^{\infty }{x}^{-0.99}dx \\ {\int }_1^{\infty }\frac{1}{x^p}dx={\int }_1^{\infty }\frac{1}{x^{{\displaystyle 0.99}}}dx \\ p=0.99\&0.99<1 \end{gathered}$$

The Fundamental Theorem of Calculus tells us that the improper integral on $[1,\infty )$diverges.

$$\begin{gathered} {\int }_1^{\infty }{x}^{-0.99}dx=\underset{b\to \infty }{\lim }{\int }_1^b{x}^{-0.99}dx \\ =\underset{b\to \infty }{\lim }{\left[100x^{0.01}\right]}_1^b \\ =\underset{b\to \infty }{\lim }\left(100b^{0.01}-100\right) \\ =\infty \end{gathered}$$