# Math Help - Integration:area between curve

1. ## Integration:area between curve

area between curves f(x)= x^2 and g(x)=-x^2 +1

2. $x^2 = -x^2 + 1$
$x = \pm \sqrt{\frac{1}{2}}$
$\int_{\sqrt{\frac{-1}{2}}}^{\sqrt{\frac{1}{2}}} [(-x^2 + 1) - (x^2)] dx =$

3. 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 &#x28;-x&#x5e;2&#x2b;1&#x29;-&#x28;x&#x5e;2&#x29; dx - Wolfram|Alpha

Here is a picture of their graphs. You want the area inbetween them.
Plot x&#x5e;2, -x&#x5e;2&#x2b;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:

$(-x^{2}+1)-(x^{2})$ or
$-2x^{2}$

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 $-Sqrt(2)/2$ and $+Sqrt(2)/2$.

This means your integral should look like:
$\int_{-Sqrt(2)/2}^{+Sqrt(2)/2} -2x^{2} dx$

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 &#x28;-x&#x5e;2&#x2b;1&#x29;-&#x28;x&#x5e;2&#x29; dx from -Sqrt&#x28;2&#x29;&#x2f;2 to Sqrt&#x28;2&#x29;&#x2f;2 - Wolfram|Alpha

4. How come the limits is sqrt 1/2
and the other time its sqrt2/2?

I am getting limits of sqrt(1/2)
and a final answer of 1.65 area?

5. Oh, I see, this confused me for a mintue, but check this out:

$Sqrt(.5)=.7071037812$
$Sqrt(2)/2=.7071067812$

They actually turn out the be the same thing.

We are both right .

Okay, so we want $(-x^{2}+1)-(x^2)$ because we want the top curve minus the bottom curve.
$-2x^{2}+1$

Lets integrate this without the limits first...
$\int-2x^{2}+1$
$\dfrac{-2}{3}x^{3}+x+c$
***Please not that this has a $\dfrac{-2}{3}$ NOT $\dfrac{+2}{3}$. In my first post, i overlooked that you missed the negative sign.
Now put in the limits of integration: $\sqrt{.5}$ and $-\sqrt{.5}$....

I get .9428....or the exact answer is: $\dfrac{2}{3}\sqrt{2}$

Hopefully, I got everything this time?