It is proved that the continuous function f (x) satisfies: ∫ (0 to 1) f (TX) DT = f (x) + xsinx

It is proved that the continuous function f (x) satisfies: ∫ (0 to 1) f (TX) DT = f (x) + xsinx

Let u = tx,du = x dtL = ∫(0~1) ƒ[tx] dt= [1/x]∫(0~x) ƒ[u] du = ƒ[x] + xsinx∫(0~x) ƒ[u] du = xƒ[x] + x²sinxƒ[x] = xƒ'[x] + ƒ[x] + 2xsinx + x²cosx...