## anonymous 5 years ago area between two graphs. y=2 and y=x^2-2

1. anonymous

You know how to integrate?

2. anonymous

yes

3. anonymous

Ok. So to find the area under the function you just integrate it right?

4. anonymous

set the two equations equal to find the bounds

5. anonymous

Since the area under the top function is the area between them and the area under the bottom function. The area between them would be Area under top function - Area under bottom function

6. anonymous

So find the x values for which the two functions are the same, then integrate each function between those bounds, and subtract the resulting areas. Sometimes it's easiest to combine the two integrals and simplify before evaluating the integral, but sometimes it's not.

7. anonymous

would it be 10.66 repeating?

8. anonymous

yes

9. anonymous

Actually, since your parabola is under the y=0 line for a bit this method won't quite work. You'll need to break up the integrals. You'll have $Area = \int\limits_{-2}^{-\sqrt{2}}[2 - (x^2-2)]dx - \int\limits_{-\sqrt{2}}^{\sqrt{2}}(x^2-2)dx + \int\limits_{\sqrt{2}}^{2}[2 - (x^2-2)]$

10. anonymous

thats nonsence... just integrate $\int\limits\limits_{-2}^{2}2-x ^{2}+2 dx=\int\limits\limits_{-2}^{2}4-x ^{2} dx$

11. anonymous

That doesn't work. Because part of that integral will be negative area.

12. anonymous

no, since you are adding 2 to the parabola in fact. it never crosses x-axis

13. amistre64

for practice, let me see what I can mess up :) I agree that the interval is [-2,2], but since we have mirror images across the yaxis, we can just do [0,2] and double it. y = 2 is just a box that is 2(2 by 2) = 8 (S)x^2-2 dx -> x^3/3 -2x |0,2 = 2[(8/3) -6] = 16/3 -36/3 = -20/3 8 --20/3 = 24/3 + 20/3 = 44/3 I would say, and this is my guess.. that the area between the 2 curves is 44/3. Now just let me know what I did wrong :)

14. anonymous

everything :) $x^{3}/3 - 2x$ from 0 to 2 is 8/3-4=8/3-12/3=-4/3 *2 => -8/3 8-(-8/3)=32/3=10+2/3=10.66666.... :D