Answer
$y=\int_{x_{0}}^x f(x) dx+y_0$
Work Step by Step
As we are given that $\dfrac{dy}{dx}=f(x)$ and $y(x_0)=y_0$
Here, we have:
$\int_{y_{0}}^y \dfrac{dy}{dx}=\int_{x_{0}}^x f(x) dx$
This gives that
$y-y_0=\int_{x_{0}}^x f(x) dx$
$y=\int_{x_{0}}^x f(x) dx+y_0$
