A curve y=f(x) which passes through (4,0) satisfies the differential equation xdy+2ydx=x(x−3)dx . The area bounded by y=f(x) and line y=x (in square unit) is

xdy+2ydx=x(x−3)dx⇒x2dy+2xydx=x3−3x2dx⇒dx2y=x3−3x2dx⇒yx2=x44−x3+cCurve passes through the point (4,0). ∴c=0 Therefore, curve is y=x24−x and point of intersection of y=x and this curve is x=x24−x   ⇒8x-x2=0  ⇒x=0, x=8 Required area =∫08 x−x24−xdx=x2−x31208=64−23×64=643