Find the area of the region bounded by the graphs of fxx24x1

Solution

Below procedure is correct. change f(x) and g(x) to get correct result .ThanQ f(x) = 2x g(x) = 8 - x^2 To determine the area of the region bounded by these two curves, you must determine where they intersect. You find these points by equating them to each other and then solving for x. f(x) = g(x) 2x = 8 - x^2 x^2 + 2x - 8 = 0 (x + 4)(x - 2) = 0 x = { -4, 2 } So your bounds are from -4 to 2. Now, you want to determine which of f(x) and g(x) is higher on that interval, because the area is going to be A = Integral (a to b, {higher curve} - {lower curve} dx ) To determine which of f(x) or g(x) is higher on that interval, choose any value in the interval (Let\'s choose x = 0), plug into both functions, and see which is higher. f(0) = 2(0) = 0 g(0) = 8 - 0^2 = 8 Clearly, g(x) is higher, making our formula A = Integral (-4 to 2, [g(x) - f(x)] dx ) A = Integral (-4 to 2, ( [8 - x^2] - 2x ) dx ) A = Integral (-4 to 2, (8 - x^2 - 2x) dx ) A = [ 8x - (1/3)x^3 - x^2 ] {evaluated from -4 to 2} A = [ 8(2) - (1/3)2^3 - 2^2] - [ 8(-4) - (1/3)(-4)^3 - (-4)^2 ] A = [ 16 - (8/3) - 4 ] - [ -32 - (1/3)(-64) - 16 ] A = [ 12 - (8/3) ] - [ -48 + (64/3) A = 12 - (8/3) + 48 - (64/3) A = 60 - (72/3) A = 60 - 24 A = 36