AP Calc– help, please!

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483

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What is the area of the region between y = 2x^2 and y = 12 - x^2?

A. 0

B. 16

C. 32

D. 48

Guest Apr 25, 2019

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3⁺⁰ Answers

+128460

+1

Find the intersection points

2x^2 = 12 - x^2

3x^2 = 12

x^2 = 4

x = - 2 and x = 2

So we have, using symmetry

2 2 2

2 * ∫ 12 - x^2 - 2x^2 dx = 2 ∫ 12 - 3x^2 dx = 2 [ 12x - x^3] = 2 [ 12(2) - (2)^3 ] =

0 0 0

2 [ 24 - 8] =

2 * 16 =

32 units^2

CPhill Apr 25, 2019

0

I'm sorry, I'm just a bit confused. When I did it, I got 16. Where does the other 2 come from? (Also, thank you very much for your help!)

Guest Apr 25, 2019

+128460

+2

We are using the fact that both functions are symmetric about the y axis

So

2 2

∫ 12 - 3x^2 dx is the same as 2 * ∫ 12 - 3x^2 dx

-2 0

CPhill Apr 25, 2019

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