> [!NOTE] Theorem > The solution to the [[Initial Value Problem for Scalar Ordinary Differential Equation|IVP]] $\frac{d}{dt}x(t)=f(t), \quad t\in (\alpha,\beta)$subject to $x(t_{0})=x_{0}$ is given by $x(t)=x_{0}+\int_{t_{0}}^{t} f(\tau) \, d\tau $ Proof: As $x_{0}$ does not depend on $t,$ it satisfies the differential equation because $\frac{d}{dt} x(t)= \frac{d}{dt} \left( \int_{t_{0}}^{t} f(\tau) \, d\tau \right)=f(t)$by [[Fundamental theorem of calculus]]. Also it satisfies the initial condition because $x(t_{0})=x_{0} + \underbrace{\int_{t_{0}}^{t_{0}} f(\tau) \, d\tau}_{=0} = x_{0} $