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

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### 2 Answers

Let f(x) = x^3 - x

We need to find the integral of f(x) on the interval [ -1, 1].

Let us assume that F(x) = integral f(x).

Then the integral is :

A = F(1) - F(-1).

Let us integrate f(x).

==> F(x) = intg x^3 - x dx

= intg x^3 dx - intg x dx

= x^4/4 - x^2/2

==> F(x) = (1/4)x^4 - (1/2)x^2.

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

= 1/4 - 1/2 = -1/4

==>** F(-1) = -1/4.**

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

==>** F(1) = -1/4**

==> A = F(-1) - F(1) = -1/4 - (-1/4) = 0

**Then the integral of f(x) from x= -1 to x= 1 is 0.**

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.