## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 8 - Techniques of Integration - 8.8 Probability and Integration - Exercises - Page 448: 7

#### Answer

$p(x)$ is a probability density function on $[1, \infty)$. and $\mu=\dfrac{3}{2}$

#### Work Step by Step

We have: $\int_{1}^{\infty} 3x^{-4} \ dx=\lim\limits_{a \to \infty}\int_{1}^{a} 3x^{-4} \ dx\\=(-1) \lim\limits_{a \to \infty} \dfrac{1}{a^3}-1\\=(-1) (-1)\\=1$ This implies that $p(x)$ is a probability density function on $[1, \infty)$. The mean value is: $\mu=\int_{1}^{\infty} (x) 3x^{-4} \ dx \\=\lim\limits_{a \to \infty}\int_{1}^{a} 3x^{-3} \ dx\\=\lim\limits_{a \to \infty} [\dfrac{-3}{2}x^{-2}]_1^a\\=\dfrac{-3}{2} \lim\limits_{a \to \infty} \dfrac{1-a^2}{a^2} \\=\dfrac{3}{2}$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.