find the area of the region enclosed by the graphsof y x32x

Solution

y = 2x + 15 and y = x^2 Equate both to get limits of integration x^2 = 2x + 15 x^2 - 2x - 15 = 0 (x - 5)(x + 3) = 0 x = -3 and 5 The graph of 2x + 15 will be on top of the y = x^2 so required area = ?( 2x + 15 - x^2 ) dx from [ - 3 to 5 ] = x^2 + 15x - (1/3)x^3 from -3 to 5 = [ (25 + 75 - 125 /3) - (9 - 45 + 27 / 3 ) ] = [ 100 + 36 - 152/3 ] = 256/3 = 85.33 sq.units