Find the area enclosed by the curves x = - 2, x = 3, y = 10x + 4, and y = 2x - 1

Question

Find the area enclosed by the curves x = - 2, x = 3, y = 10x + 4, and y = 2x - 18.

Explanation / Answer

a. The graphs of y = x + 1 and y = 9 - x² intercept at x1 and x2, roots of equation: x + 1 = 9 - x² x² + x - 8 = 0 x1 = (-1 - v33)/2 < -1; x2 = (-1 + v33)/2 > 2 Both x1 and x2 are outside of the interval [-1,2]. Over this interval, we always get (9 - x²) > (x + 1) The area of the region enclosed by the given curves and the verticals x = -1 and x = 2 is the integral of (9 - x^2) - (x + 1) = -x² - x + 8 taken over [-1,2] . . . 2 A = ? (-x² - x + 8)dx . . .-1 . .= [-(2³/3) - (2²/2) +8(2)] - [-(1³/3) -(1²/2) + 8(1)] . .= -(8/3) - 2 + 16 + (1/3) + (1/2) - 8 . .= 6 - (7/3) A = 11/3 b. The graphs of y = v(x + 3) and y = (x + 3) / 2 intercept at points with x-coordinate x1, x2, roots of equation v(x + 3) = (x + 3) / 2 2v(x + 3) = (x + 3) 4(x + 3) = (x + 3)² (x + 3)² - 4(x + 3) = 0 (x + 3)(x - 1) = 0 x1 = -3, x2 = 1 . . . 1 A = ? |v(x + 3) - (x + 3)/2| dx . . .-3 Since, over interval [-3,1], v(x + 3) > (x + 3)/2, we get . . . 1. . . . . . . . . .1 A = ? v(x + 3)dx - ?(x + 3)/2] dx . . .-3. . . . . . . . . -3 An anti-derivative of f(x) = v(x + 3) is F(x) = (2/3)(x + 3)v(x + 3) F(1) = (2/3)(1 + 3)v(1 + 3) = 16/3 F(-3)= (2/3)(-3 + 3)v(-3 + 3) = 0 An anti-derivative of g(x) = (x + 3)/2 is G(x) = x²/4 + 3x/2 G(1) = 1²/4 + 3(1)/2 = 7/4 G(-3)= (-3)²/4 + 3(-3)/2 = -9/4 Hence . . . 1. . . . . . . . . .1 A = ? v(x + 3)dx - ?(x + 3)] dx = [F(1) - F(-3)] - [G(1) - G(-3)] . . .-3. . . . . . . . . -3 A = [(16/3) - 0] - [(7/4) - (-9/4)] = 16/3 - 4 A = 4/3 u can take help from this

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