Find the Area of the Region?

Question

I am trying to do as much homework as I can today and so if you see a lot of questions from me I am not trying to have anyone do my homework just trying to understand it, with that being said here is my problem.

y=4x^2, y=x^2 + 2

Now I know what the graph looks like. You have 4x^2 which is a skinny parobala and then x2+2 which is a normal porabala starting at y=2.

I know you need to find the intercepts and such which you do by setting them equal 4x^2=x^2+2

After doing that I get to the point of 3x^2+2 but am lost as to where I go from there! Any help would be great!

I am sorry I had 3x^2-2 at the end.

The answer is not 2. Now if you get + or - the square root of 2/3 you would then itegrate that into (x2+2)-(4x2)?

Well I know we are not looking for points on this problem. I know it should be a final value I think your supposed to find the the zeroes and then intregrate?

The answer wasn't 2 either.

Answer 1

Curves intersect when:-
x² + 2 = 4x²
3x² = 2
x = ± √(2 / 3)
Curves are symmetrical about the y axis.
Area = 2 ∫ - 3.x² + 2 . dx between limits 0 and √(2 / 3)
A = 2 [- x³ + 2x ] between limits as above
Inserting limits gives:-
A = 2 [(4/3).√2/√3]
A = (8/3).(√2 / √3)
Not much fun typing this out but hopefully method is clear!

Answer 2

2times integral from 0 to 1 of (x^2+2-4x^2)dx=?

Check your answer , is it 2 ?

Answer 3

3x^2 - 2 = 0
3x^2 = 2
x^2 = 2/3
x = + / - sqrt(2/3)

Answer 4

umm your right when you set them equal, but wrong after that
4x^2=x^2+2
3x^2-2=0
3x^2+0x-2=0
quadratic formula:

x= -0(+/-)SQRT[0-4(-6)]/6
=+/-SQRT(24)/6
x=+/-SQRT(6)/3
x=0.816 OR x=-0.816
4 points: (0.816, 8), (0.816, 4), (-0.816, 8), (-0.816, 4)