How do you evaluate the integral #int 1/x^2 dx# from 1 to #oo#?
1 Answer
Aug 1, 2016
Explanation:
We have:
#int_1^oo1/x^2dx=int_1^oox^-2dx=lim_(trarroo)int_1^tx^-2dx#
Using the typical integration rule:
#=lim_(trarroo)[-1/x]_1^t#
#=lim_(trarroo)(-1/t-(-1/1))#
#=lim_(trarroo)(1-1/t)#
Note that as
#=1-0#
#=1#