# OpenStax Errata

From Chapter 6.1 EXAMPLE 6.9

This integral is solved by a change of variables, and the solution contains a factor-of-ten error.

$J=\int_0^{50}\frac{vdv}{.001v^2+9.8}$

$\text{Let }u=.001v^2+9.8$

$du=.002\,vdv$

$vdv=\frac{du}{.002} = 5\times 10^2 du \;( \text{not }5\times 10^3)$

$J=\left(5\times 10^2\right)\int\frac{du}{u}=\left(5\times 10^2\right) \left.\ln\left(.001v^2+9.8\right)\right|_0^{50}\approx 114\text{ m}$

The book's numerical value is correct, but one step in the equation is wrong. This error was pointed out to me by a student at Wright State University Lake Campus.--Guy vandegrift (talk) 17:56, 13 February 2018 (UTC)