Hi,
I have this question here that I'm not sure about.
Any help is appreciated. Evaluate the integral from 1 to 3 of dx/(x-2). This seems like it should be really simple- is it 0?
∫dx/(x-2) = ln(x-2)
first off, ln(x-2) is not defined for x < 2. So, some like to evaluate it as
∫[1,3] dx/(x-2) = ln|x-2| [1,3] = ln|3-2|-ln|1-2| = ln1 - ln1 = 0-0 = 0
That is because the graph is symmetric about (1,0) so the upper and lower areas cancel out.