Re: Integration

cooksec@nationwide.com
Fri, 23 Apr 1999 10:36:12 -0400

Use the substitution... u=10+x (which then gives x=u-10 and du=dx)

This substitution leaves you with the integrand... [(u-10)du]/u^2 or
[du/u - 10du/u^2]

This is easily integrable, but be sure to switch your limits of integration from
0 and 5, to 10 and 15.

Without simplifying, I got as an answer... ln(15) - ln(10) + 10/15 -
10/10

This can simplify to... ln(3/2) - 1/3 = .072

Chris