Find this integral?

Question

Find the integral of 3/x^2 dx from -1 to 1. 

 

Answer: nonexistent

 

Please show all your work. 

Answer 1

The function y=3/x² has discontinuity at x=0. Moreover, it tends to infinity as x→0. Therefore, we must break integral into two parts: from -1 to -ε and from ε to 1 and see whether the limit of the values of those integrals exists as ε→0.  _-1∫^-ε 3/x² dx=-(3/x)_-1^-ε=3/ε-3; _ε∫¹ 3/x² dx=-(3/x)_ε¹=-3+3/ε;

 

The whole integral is the sum of two integrals, that is, 6/ε-6;

Now it is clear that the limit of this expression as ε→0 does not exist. Therefore, the whole integral does not exist either.

 

Answer 2

f(x) = 3/x^2 has a vertical asymptote at x = 0.

 

In fact, as x → 0^±, f → +∞.

 

So you can't integrate f around 0.

Answer 3

The function 3/x² is symmetric with respect to chnange in sign: x by - x. This means the integral 

                        1                   1

                        ∫(3/x²)dx = 2∫(3.x²)dx

                      -1                   0

 

Because the function converges when x → 0, this integral → ∞.

Answer 4

   1                        1   

∫    3dx /x^2 = - 3 l    =( - 3/1 )  - ( -3/ -1 ) = -3 -3 = -6

 -1                    X  l-1