Question

The finite region bounded by y=12-2x^2 and y=x^2-8. (This is for area between curves, Calculus)

Answer 1

≈ 68.8530

Explanation:

You want the area between the curves y = 12 -2x² and y = x² -8.

Area

The area will be the integral of the difference between the curves, between the points where the curves intersect.

h(x) = (12 -2x²) -(x² -8) = 20 -3x² . . . . . height of a differential of area

The point where the curves intersect has an x-value that makes h(x) = 0. We choose to call the positive value 'a'. The curves also intersect at x=-a.

h(a) = 0

20 -3a² = 0

a² = 20/3

a = √(20/3) = (2/3)√15

So, the integral is ...

`\displaystyle A=\int_(-a)^a{(20-3x^2)}\,dx=\left(20x-(3x^3)/(3)\right)_(-a)^a\\\\A=2a(20-a^2) = (4)/(3)√(15)\cdot\left(20-(20)/(3)\right)\\\\A=\boxed{(160√(15))/(9)\approx68.8530}`

The area between the curves is about 68.8530 square units.