How do you evaluate the integral #int 1/xdx# from [-3,-2]?
1 Answer
Feb 23, 2017
Note that
So, when we apply the bounds, we see that:
#int_(-3)^(-2)1/xdx=[lnabsx]_(-3)^(-2)#
Evaluating this now:
#=lnabs(-2)-lnabs(-3)=ln2-ln3=ln(2/3)#
