YES!(WELL THIS IS AN EDIT AND ACTUALLY NO, ITS OFF BY A NEGATIVE SIGN...SEE SECOND POST) This is the answer except you might want to include the limits. Because this is a closed region you can find the area.
You set everything up right just like this: you might want to include limits depending on if you questions wants the actual area or the expresssion for the area.
integral of (-x^2+1)-(x^2) dx - Wolfram|Alpha
Here is a picture of their graphs. You want the area inbetween them.
Plot x^2, -x^2+1 - Wolfram|Alpha
-x^2+1 has the graph on top of the other for the region we are looking for. So the expression that we find the integral over is would be:
or
The limits of integral are from where the two graphs intersect. Set the two equations equal to eachother to find what x values they have in common....The intersection points are and .
This means your integral should look like:
You should get (2/3)Sqrt(2) as the area between the curves. (or just substitute in the limits into your expression that you already have, these are the same things!!).
integral of (-x^2+1)-(x^2) dx from -Sqrt(2)/2 to Sqrt(2)/2 - Wolfram|Alpha
Oh, I see, this confused me for a mintue, but check this out:
They actually turn out the be the same thing.
We are both right .
Okay, so we want because we want the top curve minus the bottom curve.
Lets integrate this without the limits first...
***Please not that this has a NOT . In my first post, i overlooked that you missed the negative sign.
Now put in the limits of integration: and ....
I get .9428....or the exact answer is:
Hopefully, I got everything this time?