Constant of Integration — Question 1

Question 1 - Solution

Since $F$ and $G$ are both antiderivatives of $f$ on $I$, we have $$F'(x)=f(x) \quad\text{and}\quad G'(x)=f(x) \quad\text{for all }x\in I.$$

Consider the function $$H(x)=F(x)-G(x).$$

Differentiate $H(x)$: $$H'(x)=F'(x)-G'(x)=f(x)-f(x)=0 \quad\text{for all }x\in I.$$

Thus, the derivative of $H$ is zero on the interval $I$.

From the result that a function with zero derivative on an interval is constant, there exists a real number $C$ such that $$H(x)=C \quad\text{for all }x\in I.$$

Substituting back for $H(x)$ gives $$F(x)-G(x)=C.$$

Rewriting, $$F(x)=G(x)+C.$$

$$\boxed{F(x)=G(x)+C}$$