x f'(x)=2 If f′(x)=2f'(x) = 2f′(x)=2 and f(0)=5f(0) = 5f(0)=5, then f(x)=f(x) =f(x)=:A2x+52x + 52x+5check_circleB2x2x2xCx+5x + 5x+5D2x2+52x^2 + 52x2+5Explanationf(x)=∫2 dx=2x+Cf(x) = \int 2\,dx = 2x + Cf(x)=∫2dx=2x+C; f(0)=5f(0) = 5f(0)=5 gives C=5C = 5C=5.