A real valued differentiable function f satisfies f(x)+f(y)+2 x y=f(x+y) x, y R . Given f^n(0)=0 , then value of ₀^ 2 f( x) d x will be

Step-by-step solution

f(x)+f(y)+2 x y=f(x+y)...(i) differentiating the equation, f^ (y)+2 x=f^ (x+y) at y=0f^ (0)+2 x=f^ (x) f^ (x)=2 x Integrating both sides. f^ (x)= 2 x d x f(x)=x^2 I= ₀^ 2 f( x) d x I= ₀^ 2 ^2 x d x...(ii) I= ₀^ 2 ^2 ( 2 -x ) d x= ₀^ 2 ^2 x d x (iii) Adding Eqs. (ii) and (iii) 2l= ₀^ 2 ( ^2 x+ ^2 x ) d x= ₀^ 2 1 d x array ll & 2l= 2 & l= 4 array