Let f:(0,∞)→R and F(x)=∫0xf(t)dt. If Fx2=x2(1+x) then f(4) equals

Fx2=∫0x2f(t) dt , there fore, x2(1+x)=∫0x2f(t)dt.  Differentiating both sides w.r.t. x using Property17, we have2x+3x2=fx2·2x⇒fx2=1+(3/2)xPutting x=2 we have f(4)=1+3=4.