Finding area of a region on graph using integration.

Question

Consider the following.

y = x²

y = 12 − x

(a) Find the area of the region by integrating with respect to x.

(b) Find the area of the region by integrating with respect to y.

Any help will be very appreciated, thank you in advance!

Answer 1

(a) Limits of integration: x² = 12 - x; x² + x - 12 = 0; (x + 4)(x - 3) = 0; x = - 4 or x = 3.

Area A = ∫_-4³(12 - x - x²)dx = (12x - x²/2 - x³/3)_-4³ = (36 - 9/2 - 9) - (- 48 - 16/2 + 64/3) = 343/6;

(b) Limits of integration: -√y ≤ x ≤ √y, 0 ≤ y ≤ 9; - √y ≤ x ≤ 12 - y; 9 ≤ y ≤ 16

So, area A = ∫₀⁹2√ydy + ∫₉¹⁶(12 - y + √y)dy = 4/3y^3/2₀⁹ + (12y - y²/2 + 2/3y^3/2)₉¹⁶ = 4/3·27 + (192 - 128 +

2/3·64) - (108 - 81/2 + 2/3·27) = 343/6

Answer 2

INTEGRATING WITH RESPECT TO X

Area = ∫^{x= 3}_x=-4 [ 12 - x - x² ] d x = [ 12x -x²/ 2 - x³/3 ] |_-4³ = 343 / 6 square units.

INTEGRATING WITH RESPECT TO Y

Area = ∫^y=9_y=0 [ √y - ( - √y )] d y + ∫^y=16_y=9 [ 12 -y - ( - √y )] d y = 343 / 6 square units.