Schematically plot region
Area A = A1 + A2
Parts
A 1 = ∫ x 1 x 2 ∫ 11 / 16 x 2 3 − 2 x d y d x A1 = \int_{x1}^{x2}\int_{11/16x^2}^{3-2x}dydx A 1 = ∫ x 1 x 2 ∫ 11/16 x 2 3 − 2 x d y d x
A 2 = ∫ x 2 0 ∫ 11 / 16 x 2 x 2 d y d x A2 = \int_{x2}^{0}\int_{11/16x^2}^{x^2}dydx A 2 = ∫ x 2 0 ∫ 11/16 x 2 x 2 d y d x
Integration boundaries
x 1 : 3 − 2 x 1 = 11 / 16 x 1 2 , x 1 ≤ 0 x 1 = − 4 x1:\\
3 - 2x_1 = 11/16x_1^2, x_1 \le 0\\
x_1 = -4 x 1 : 3 − 2 x 1 = 11/16 x 1 2 , x 1 ≤ 0 x 1 = − 4
x 2 : 3 − 2 x 2 = x 2 2 , x 2 ≤ 0 x 2 = − 3 x2:\\
3-2x_2 = x_2^2, x_2 \le 0\\
x_2 = -3 x 2 : 3 − 2 x 2 = x 2 2 , x 2 ≤ 0 x 2 = − 3
Area parts
A 1 = ∫ − 4 − 3 ∫ 11 / 16 x 2 3 − 2 x d y d x = ∫ − 4 − 3 3 − 2 x − 11 / 16 x 2 d x = [ 3 x − x 2 − 11 / 48 x 3 ] − 4 − 3 = 3 ∗ ( − 3 + 4 ) − ( 9 − 16 ) − 11 / 48 ∗ ( − 27 + 64 ) ≈ 1.5208 A1 = \int_{-4}^{-3}\int_{11/16x^2}^{3-2x}dydx\\
= \int_{-4}^{-3} 3 - 2x - 11/16x^2dx\\
= \Big[ 3x - x^2 - 11/48x^3 \Big]_{-4}^{-3}\\
= 3*(-3 + 4) - (9 - 16) - 11/48*(-27 + 64) \approx 1.5208 A 1 = ∫ − 4 − 3 ∫ 11/16 x 2 3 − 2 x d y d x = ∫ − 4 − 3 3 − 2 x − 11/16 x 2 d x = [ 3 x − x 2 − 11/48 x 3 ] − 4 − 3 = 3 ∗ ( − 3 + 4 ) − ( 9 − 16 ) − 11/48 ∗ ( − 27 + 64 ) ≈ 1.5208
A 2 = ∫ − 3 0 ∫ 11 / 16 x 2 x 2 d y d x = ∫ − 3 0 x 2 − 11 / 16 x 2 d x = ∫ − 3 0 5 / 16 x 2 d x = 5 / 48 x 3 ∣ − 3 0 = 5 / 48 ∗ ( 0 + 27 ) = 2.8125 A2 = \int_{-3}^{0}\int_{11/16x^2}^{x^2}dydx\\
= \int_{-3}^{0} x^2 - 11/16x^2dx=\int_{-3}^{0} 5/16x^2dx\\
= 5/48x^3\Big|_{-3}^{0}\\
= 5/48*(0 + 27) = 2.8125 A 2 = ∫ − 3 0 ∫ 11/16 x 2 x 2 d y d x = ∫ − 3 0 x 2 − 11/16 x 2 d x = ∫ − 3 0 5/16 x 2 d x = 5/48 x 3 ∣ ∣ − 3 0 = 5/48 ∗ ( 0 + 27 ) = 2.8125
Area
A = A 1 + A 2 ≈ 1.5208 + 2.8125 = 4.3333 A = A1 + A2 \approx1.5208 + 2.8125 = 4.3333 A = A 1 + A 2 ≈ 1.5208 + 2.8125 = 4.3333
#340153 on Dec 2023