Chapter 6.4, Problem 31E

To determine

To evaluate the given integral.

Answer to Problem 31E

  ${\displaystyle \underset{1}{\overset{32}{\int }}{x}^{-\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}dx}=\frac{5}{2}$

Explanation of Solution

Given:

  ${\displaystyle \underset{1}{\overset{32}{\int }}{x}^{-\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}dx}$

Calculation:

  $\begin{array}{l}{\displaystyle \underset{1}{\overset{32}{\int }}{x}^{-\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}dx}={\left[\frac{x^{-\frac{6}{5}+1}}{-\frac{6}{5}+1}\right]}_1^{32}\\ ={\left[\frac{x^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}}{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}\right]}_1^{32}\\ =-5\left[{\left(32\right)}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}-{\left(1\right)}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}\right]\\ =-5\left[{\left(2^5\right)}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}-1\right]\\ =-5\left[\frac{1}{2}-1\right]\\ =\frac{5}{2}\end{array}$