evaluate the integral of x^3-x if x=-1 and x=1

We'll apply the fundamental theorem of
calculus:

Int f(x)dx = F(b) - F(a) for x = a to x  =
b

In this case a = -1 and b =
1

We recall the property of additivity of
Integrals:

Int (x^3-x)dx = Int x^3dx - Int
xdx

Int (x^3-x)dx = x^4/4 - x^2/2 + C (we've applied the
formulas from the table of elementary indefinite
integrals)

But F(x) = x^4/4 - x^2/2 +
C

We'll calculate F(1) and
F(-1):

F(1) = 1/4 - 1/2

F(-1)
= 1/4 - 1/2

F(1) - F(-1) =  1/4 - 1/2 - 1/4 +
1/2

We'll eliminate like terms and we'll
get:

Int (x^3-x)dx =
0

Note: The integral of odd functions, for
the symmetric limits, x = -a to x = a, is cancelling.