# Thread: Area under the curve integration

1. ## Area under the curve integration

Find the total area enclosed by the curves and
I drew out the graph I just am not sure how to do this would problem. Or how to integrate absolute values.
Thanks to anyone who can help!

2. Originally Posted by Casas4
Find the total area enclosed by the curves and
I drew out the graph I just am not sure how to do this would problem. Or how to integrate absolute values.
Thanks to anyone who can help!

$\displaystyle A = 2\int\limits_0^2 {\left( {x - \left( {{x^2} - 2} \right)} \right)dx} = \ldots$

Look a graph of these function and remember the definition of the module.

3. Ok that makes since but how come we arent taking into consideration the negative side of the graph (or the left side)?

4. Originally Posted by Casas4
Ok that makes since but how come we arent taking into consideration the negative side of the graph (or the left side)?
We take into account the left side of the graph, and hence the integral is multiplied by two (because figure, formed by the intersection of these graphs, has symmetry about the vertical axis).
Do you understand this?

5. Hello, Casas4!

If, as you say, you drew the graph,
. . I don't understand your difficulty.

Find the total area enclosed by the curves: .$\displaystyle \begin{array}{c}y \:=\:|x| \\ y \:=\:x^2-2 \end{array}$
Code:
                |       (2,2)
*         |         *
::*       |       *::
:::*     |     *:::
*::::*   |   *::::*
::::::* | *::::::
- - *:-:-:-:*:-:-:-:* - - -
*::::::|::::::*
*:::|:::*
*
|

Due to the symmetry, the area is: .$\displaystyle 2 \times \int^2_0\bigg[x - (x^2-2)\bigg]\,dx$