Application of definite integral

[cmnalo]

f(x) =x^3
g(x) = x

x^3 =x
X^3 -x =0
x(x^2-1)=0
x(x-1)(x+1) =0
x=1
x=-1

∫ [(x^3) -(x)]dx [-1,1]
∫ x^3 dx - ∫ xdx
1/4 x^4 -1/2x^2
1/4(1)^4 -1/2(-1)^2
1/4 -1/2 =-1/4

My book says the answer is 1/2. Where did I go wrong?

 

[stapel]

Where did I go wrong?

I dunno. What are you doing with f and g?

Please reply with the full text of the exercise, including the instructions. Thank you.

Eliz.

 

[galactus]

It's an odd function and symmetric about the y-axis, therefore,

\(\displaystyle \L\\2\int_{0}^{1}(x-x^{3})dx\)

or

\(\displaystyle \L\\2\int_{-1}^{0}(x^{3}-x)dx\)

Graph it and you'll see why.

 

[cmnalo]

stapel-
I'm trying to find the area of the region bounded by the graphs of the two functions.

A= ∫ [f(x) -g(x)]dx [a,b]

 

[soroban]

Hello, cmnalo!

Find the area between: $\displaystyle f(x) \,=\,x^3$ and $\displaystyle g(x) \,=\, x$

Did you make a sketch? . . .

                      |
                      |             *(1,1)
                      |           *::
                *     |        *::::
             *:::::*  |     *::::::*
           *:::::::::*|  *:::::::::
      ----*:-:-:-:-:-:*:-:-:-:-:-:*----
          :::::::::*  |*:::::::::*
         *::::::*     |  *:::::*
         ::::*        |     *
        ::*           |
 (-1,-1)*             |