the integral in this exercise converges. evaluate the integral without using a table.\n int_{1}^{infty}…

the integral in this exercise converges. evaluate the integral without using a table.\n int_{1}^{infty} \frac{1}{x^{1.0001}} dx \n int_{1}^{infty} \frac{1}{x^{1.0001}} dx =

the integral in this exercise converges. evaluate the integral without using a table.\n int_{1}^{infty} \frac{1}{x^{1.0001}} dx \n int_{1}^{infty} \frac{1}{x^{1.0001}} dx =

Answer

Explanation:

Step1: Use the power - rule for integration

The power - rule for integration is (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C) ((n\neq - 1)). For the integral (\int\frac{1}{x^{1.0001}}dx=\int x^{-1.0001}dx), where (n=-1.0001). Then (\int x^{-1.0001}dx=\frac{x^{-1.0001 + 1}}{-1.0001+1}+C=\frac{x^{-0.0001}}{-0.0001}+C=-\frac{10000}{x^{0.0001}}+C).

Step2: Evaluate the improper integral

The improper integral (\int_{1}^{\infty}\frac{1}{x^{1.0001}}dx=\lim_{b\rightarrow\infty}\int_{1}^{b}x^{-1.0001}dx). [ \begin{align*} \lim_{b\rightarrow\infty}\int_{1}^{b}x^{-1.0001}dx&=\lim_{b\rightarrow\infty}\left[-\frac{10000}{x^{0.0001}}\right]{1}^{b}\ &=\lim{b\rightarrow\infty}\left(-\frac{10000}{b^{0.0001}}+\frac{10000}{1^{0.0001}}\right) \end{align*} ] Since (\lim_{b\rightarrow\infty}\frac{10000}{b^{0.0001}} = 0) (because for any positive (\alpha), (\lim_{x\rightarrow\infty}\frac{1}{x^{\alpha}}=0)).

Answer:

(10000)