Evaluate the integral? : int_(-3)^2 1/x^4 dx

1 Answer
Mar 5, 2017

The integral is divergent and therefore int_(-3)^2 \ 1/x^4 \ dx is undefined.

Explanation:

I recommend that you always draw a sketch to clarify what needs calculating.

graph{1/x^4 [-5, 5, -2, 10]}

The curve has a singularity at x=0 we therefore need to establish the behaviour of the integral around the singularity as follows:

int_(-3)^2 \ 1/x^4 \ dx = int_(-3)^0 \ 1/x^4 \ dx+int_0^2 \ 1/x^4 \ dx
" " = lim_(a rarr 0^-)int_(-3)^a \ 1/x^4 \ dx+ lim_(rarr o^-)int_0^b \ 1/x^4 \ dx

" " = lim_(a rarr 0^-) [-1/(3x^3) ]_-3^a

" " + lim_(b rarr 0^+) [ -1/(3x^3) ]_(b)^2

And neither of these limits exist.