Chapter 9.4, Problem 3QQ

To determine

To find: theall values of p for which the integral converges ${\displaystyle {\int }_0^1\frac{dx}{x^{p+1}}}$ .

Answer to Problem 3QQ

for all values of pthe integral converges ${\displaystyle {\int }_0^1\frac{dx}{x^{p+1}}}$ .

Explanation of Solution

Given information:

The integral is ${\displaystyle {\int }_0^1\frac{dx}{x^{p+1}}}$ .

The values are $(\text{A)}p<-1\text{ (B)}p<0\text{ (C)}p>0\text{ (D)}p>1\text{ (E)diverges for all }p$ .

Calculation: to find the values, first integrate the integral,

  $\begin{array}{l}{\displaystyle {\int }_0^1\frac{dx}{x^{p+1}}}=\underset{a\to \infty }{\lim }{\displaystyle {\int }_0^1{x}^{-p-1}}\text{ [simplify]}\\ {\displaystyle {\int }_0^1\frac{dx}{x^{p+1}}}=\underset{a\to \infty }{\lim }{\left[\frac{x^{-p-1+1}}{-p-1+1}\right]}_0^1\text{ [}{\displaystyle \int x^n=\frac{x^{n+1}}{n+1}]}\\ {\displaystyle {\int }_0^1\frac{dx}{x^{p+1}}}=\underset{a\to \infty }{\lim }{\left[\frac{x^{-p}}{-p}\right]}_0^1\\ {\displaystyle {\int }_0^1\frac{dx}{x^{p+1}}}=\underset{a\to \infty }{\lim }\left[\frac{a^{-p}}{-p}+\frac{1^{-p}}{p}\right]\text{ [substitute upper and lower limit]}\end{array}$

Form the above equation, for the integral converges for all values of p.