1. ## Integral Area

Hello fellow math members!

I can't find this area of this two function intersecion:
Y1=$\displaystyle x^3-x$
Y2=sin(pi*x)

For $\displaystyle -1\leq x\leq 1$

From what I know, I need to do y1=y2 initially, right?
But even that I did't get to resolve =/

Any help? ^^

Spoiler:
Area=(8+pi)/pi

Thanks!

2. ## Re: Integral Area

A graphical representation of the intersections of the two functions would be sufficient in my opinion, but if you want to be slightly more rigorous:

When x = -1, 0, 1 then both $\displaystyle x^3-x = x(x-1)(x+1)$ and $\displaystyle sin([pi x)$ = 0
In the interval (-1,0), $\displaystyle sin([pi x) > 0$ and $\displaystyle x^3-x < 0$ so the two functions cannot be equal.
In the interval (0,1), "" < 0 and "" > 0 so the two functions cannot be equal.

Hence the only intersection points are -1, 0, 1. then set up your integrals.

one will be from -1 to 0 with $\displaystyle sin(pi x) - (x^3-x)$ and the other will be from 0 to 1 with $\displaystyle (x^3-x) - sin(pi x)$. After solving both, add them to get your total area.

3. ## Re: Integral Area

Thanks a lot mate!
Just did it!