Evaluate an integral

[kggirl]

Can someone please help with this problem. I'm not sure what to do because there is no N in the numerator.

Integral from -1 to 1 5/x3 dx

I know that normally I would add 1 to N in the numerator and the denominator, but there is no N in the numerator in this one, so what do I need to do?

Integral from -1 to 1 5/x3 dx

would it be F[x] = 5/4?

 

[Gene]

Think of it as 5x^(-3)dx

 

[kggirl]

ok, so is this correct:

integral from -1 to 1 5x^-3 dx =
5^-2/4dx =
.04/4 =
.01

 

[Gene]

Not quite.
d(u^n) = nu^(n-1) so
Integral of u^-3 du = (-1/2u^2)
Evaluating from -1 to 1 gives
(-1/1)-(-1/1) = 0
Which we might have expected 'cause
f(x) = 1/x³
f(-x) = -f(x) so whatever you get going from -a to 0 you get the same thing with the opposite sign going from 0 to a and they cancel out. Total area = 0