How do you test the improper integral #int x^-2 dx# from #[2,oo)# and evaluate if possible?

1 Answer
mason m
Apr 13, 2017

#int_2^oox^-2dx=1/2#

Explanation:

We will find the antiderivative of #x^-2# as normal. When we "evaluate" at infinity, we will take the limit of the antiderivative at infinity instead.

Note that:

#intx^-2dx=x^-1/(-1)+C#

#color(white)(intx^-2dx)=-1/x+C#

So:

#int_2^oox^-2dx=[-1/x]_2^oo#

#color(white)(int_2^oox^-2dx)=[lim_(xrarroo)(-1/x)]-(-1/2)#

The limit approaches #0#:

#color(white)(int_2^oox^-2dx)=0-(-1/2)#

#color(white)(int_2^oox^-2dx)=1/2#

So, the area under #1/x^2# from #x=2# onwards infinitely is only #1/2#.

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