Constant of Integration — Question 2

Question 2 - Solution

Let $x_1,x_2\in I$ with $x_1<x_2$. We will show that $$F(x_1)=F(x_2).$$

Since $F'(x)=0$ for all $x\in I$, the function $F$ is differentiable on $I$ and continuous on the closed interval $[x_1,x_2]$.

By the Mean Value Theorem, there exists a number $c\in(x_1,x_2)$ such that $$F'(c)=\frac{F(x_2)-F(x_1)}{x_2-x_1}.$$

But $F'(c)=0$ by assumption, so $$0=\frac{F(x_2)-F(x_1)}{x_2-x_1}.$$

Since $x_2-x_1\neq 0$, it follows that $$F(x_2)-F(x_1)=0.$$

Thus, $$F(x_2)=F(x_1).$$

Because $x_1$ and $x_2$ were arbitrary points in $I$, the function $F$ has the same value at every point of $I$.

$$\boxed{F \text{ is constant on } I}$$