Consider three simple functions F(x)=2x, G(x)=2x-6, H(x)=2x + π/2. We can differentiate these functions with respect to x to get F’(x)=f(x)=2, G’(x)=g(x)=2, H’(x)=h(x)=2. Clearly the derivatives of these functions are all equal to 2, but the functions are not the same (a simple graph would convince us). Going backwards if we’re given u(x)=2 and we are asked to find the antiderivative of u(x) (i.e. a function U(x) such that U’(x)=u(x)) we cannot simply write that U(x)=2x since that would give us only one out of the infinite antiderivatives. Hence, we are looking for a family of functions of the form U(x)=2x+C where C is any real number. To convince ourselves this is a solution we can differentiate with respect to x which gives U’(x)=u(x)=(2x+C)’=(2x)’+(C)’=2*1+0=2.