Question

area between two graphs. y=2 and y=x^2-2

Answer 1

You know how to integrate?

Answer 2

yes

Answer 3

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

Answer 4

set the two equations equal to find the bounds

Answer 5

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

Answer 6

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.

Answer 7

would it be 10.66 repeating?

Answer 8

yes

Answer 9

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
\[Area = \int\limits_{-2}^{-\sqrt{2}}[2 - (x^2-2)]dx - \int\limits_{-\sqrt{2}}^{\sqrt{2}}(x^2-2)dx + \int\limits_{\sqrt{2}}^{2}[2 - (x^2-2)]\]

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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 :)

Answer 14

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