Constant of Integration — Question 6

Question 6

Let $f$ be continuous on an interval $I$, and fix one antiderivative $F$. Write the general antiderivative as $$y(x)=F(x)+C.$$ Show that the condition $$y(x_0)=y_0,\qquad x_0\in I,$$ uniquely determines $C$.

Question 6 - Solution

Fix an antiderivative $F$ of $f$ on $I$. Every other antiderivative has the form $y(x)=F(x)+C$.

Imposing $y(x_0)=y_0$ yields $F(x_0)+C=y_0$, so

$$\boxed{C=y_0-F(x_0).}$$

This value is unique because $F$, $x_0$, and $y_0$ are fixed.