Quote (CrackerJacker @ Aug 18 2014 12:28am)
I can rewrite it as (-1) * 1/(y-10)
with (-1) acting as a constant theres no reason I shouldn't be able to take it outside of the integral
This is true. Your book is not saying that -ln(10-y) = -ln(y-10) because this is not equivalent.
However what it is saying is that -1/(y-10) = 1/(10-y).
Your issue lies when you take the integral of -1/(y-10). The answer isnt so much -ln(y-10) as it is -ln( |y-10| )
The same applies if simplify as your book does to 1/(10-y) Then your integral evaluation is -ln( |10-y| )
Which in turn are equivalent.
Let y = 5
Case 1: -ln( | y-10 | )
-ln ( | 5 - 10 | ) = -ln ( | -5 | ) = - ln (5)
Case 2: -ln ( | 10-y | )
-ln ( | 10-5 | ) = -ln (5)
This post was edited by trojan24 on Aug 17 2014 11:43pm