Proof of Various Integral Properties — Question 9

Question 9 - Solution

Define a new function $$h(x)=f(x)-m.$$

Since $f$ is integrable on $[a,b]$ and $m$ is a constant, the function $h$ is also integrable on $[a,b]$.

From the assumption $f(x)\ge m$ for all $x\in[a,b]$, we have $$h(x)=f(x)-m\ge 0 \quad\text{for all }x\in[a,b].$$

By the positivity property of integrals, $$\int_a^b h(x)\,dx \ge 0.$$

Substitute back for $h(x)$: $$\int_a^b \bigl(f(x)-m\bigr)\,dx \ge 0.$$

Use linearity of the integral: $$\int_a^b f(x)\,dx - \int_a^b m\,dx \ge 0.$$

Since $\int_a^b m\,dx = m(b-a)$, we obtain $$\int_a^b f(x)\,dx - m(b-a) \ge 0.$$

Rearranging gives $$\int_a^b f(x)\,dx \ge m(b-a).$$

$$\boxed{\int_a^b f(x)\,dx \ge m(b-a)}$$