If the area bounded by y = x^3 + ax (where a > 0 ), the x -axis and the lines x = -2 and x = 1 is 37 4 square units, then

Step-by-step solution

Given the curve y = x^3 + ax with a > 0 , the roots of y = 0 are given by x(x^2 + a) = 0 . Since a > 0 , the only real root is x = 0 . For x [-2, 0] , y 0 , and for x [0, 1] , y 0 . The required area A is given by: A = _ -2 ⁰ -(x^3 + ax) dx + ₀¹ (x^3 + ax) dx A = - [ x^4 4 + ax^2 2 ]_ -2 ⁰ + [ x^4 4 + ax^2 2 ]₀¹ A = - ( 0 - ( 16 4 + 4a 2 ) ) + ( 1 4 + a 2 - 0 ) A = (4 + 2a) + ( 1 4 + a 2 ) A = 17 4 + 5a 2 Given that the area is 37 4 square units: 17 4 + 5a 2 = 37 4 5a 2 = 20 4 5a 2 = 5 a = 2 Answer: a = 2