Question

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

Answer 1

`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`.