Constant of Integration — Question 3

Question 3 - Solution

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

Consider the difference of the two functions: $$H(x)=F(x)-G(x).$$

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 Mean Value Theorem, a function whose derivative is zero everywhere on an interval must be constant. Therefore, there exists a real number $K$ such that $$H(x)=K \quad\text{for all }x\in I.$$

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

$$\boxed{F(x)-G(x)=\text{constant}}$$