How do you evaluate #int 1/x^2# for [-1, -1/2]?

Question

1196 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-23T02:22:57

We can use the power rule.

#int 1/x^2dx = int x^(-2)dx#

#int_a^b x^ndx = [(x^(n+1))/(n+1)]|##""_(a)^(b)#

#= [(b^(n+1))/(n+1)] - [(a^(n+1))/(n+1)]#

So, let's use this formula to do so.

#=> color(blue)(int_(-1)^("-1/2") 1/x^2dx) = int_(-1)^("-1/2") x^(-2)dx#

#= [x^(-1)/(-1)]|_(-1)^"-1/2" = ##[-1/x]##|_(-1)^"-1/2"#

#= [-1/(-"1/2")] - [-1/(-1)]#

#= 2 - 1#

#= color(blue)(1)#