A region is bounded by the curve y = x|x| and the line y = kx , where k > 0 . If the total area of this enclosed region is 72 square units, then the value of k is :

Step-by-step solution

The given curves are y = x|x| and y = kx . For x 0 , the curve is y = x^2 . The intersection with y = kx is at x = k . The area in the first quadrant is: A₁ = ₀^ k (kx - x^2) dx = [ kx^2 2 - x^3 3 ]₀^ k = k^3 2 - k^3 3 = k^3 6 For x The area in the third quadrant is: A₂ = _ -k ⁰ (-x^2 - kx) dx = [ - x^3 3 - kx^2 2 ]_ -k ⁰ = 0 - ( k^3 3 - k^3 2 ) = k^3 6 The total enclosed area is A = A₁ + A₂ = k^3 6 + k^3 6 = k^3 3 . Given that the total area is 72 square units: k^3 3 = 72 k^3 = 216 k = 6 . Answer: 6