Answer to Question #336153 in Calculus for Quân Jason

Question

Find the area of the region bounded by y= 11x²/16, y=x², y=3-2x, satisfying x≤0

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Answer 1

Schematically plot region

Area A = A1 + A2

Parts

$A1 = \int_{x1}^{x2}\int_{11/16x^2}^{3-2x}dydx$

$A2 = \int_{x2}^{0}\int_{11/16x^2}^{x^2}dydx$

Integration boundaries

$x1:\\ 3 - 2x_1 = 11/16x_1^2, x_1 \le 0\\ x_1 = -4$

$x2:\\ 3-2x_2 = x_2^2, x_2 \le 0\\ x_2 = -3$

Area parts

$A1 = \int_{-4}^{-3}\int_{11/16x^2}^{3-2x}dydx\\ = \int_{-4}^{-3} 3 - 2x - 11/16x^2dx\\ = \Big[ 3x - x^2 - 11/48x^3 \Big]_{-4}^{-3}\\ = 3*(-3 + 4) - (9 - 16) - 11/48*(-27 + 64) \approx 1.5208$

$A2 = \int_{-3}^{0}\int_{11/16x^2}^{x^2}dydx\\ = \int_{-3}^{0} x^2 - 11/16x^2dx=\int_{-3}^{0} 5/16x^2dx\\ = 5/48x^3\Big|_{-3}^{0}\\ = 5/48*(0 + 27) = 2.8125$

Area

$A = A1 + A2 \approx1.5208 + 2.8125 = 4.3333$