Proof of Various Integral Properties — Question 6

Question 6

Assume that $f$ and $g$ are integrable on $[a,b]$ and that $$f(x)\le g(x) \quad\text{for all }x\in[a,b].$$ Prove that $$\int_a^b f(x)\,dx \le \int_a^b g(x)\,dx.$$

Question 6 - Solution

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

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

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

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

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

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

Rearranging gives $$\int_a^b f(x)\,dx \le \int_a^b g(x)\,dx.$$

$$\boxed{\int_a^b f(x)\,dx \le \int_a^b g(x)\,dx}$$