Proof of Various Integral Properties — Question 7

Question 7

Assume that $f$ is integrable on $[a,b]$. Prove that $$\left|\int_a^b f(x)\,dx\right| \le \int_a^b |f(x)|\,dx.$$

Question 7 - Solution

We begin with a basic inequality involving absolute values.

For all real numbers $u$, we have $$-|u|\le u \le |u|.$$

Apply this inequality to $f(x)$: $$-|f(x)| \le f(x) \le |f(x)| \quad\text{for all }x\in[a,b].$$

Integrate each part of the inequality over $[a,b]$: $$\int_a^b -|f(x)|\,dx \le \int_a^b f(x)\,dx \le \int_a^b |f(x)|\,dx.$$

Simplify the left-hand integral: $$-\int_a^b |f(x)|\,dx \le \int_a^b f(x)\,dx \le \int_a^b |f(x)|\,dx.$$

This shows that the value of $\int_a^b f(x)\,dx$ lies between $-\int_a^b |f(x)|\,dx$ and $\int_a^b |f(x)|\,dx$.

By definition of absolute value, this implies $$\left|\int_a^b f(x)\,dx\right| \le \int_a^b |f(x)|\,dx.$$

$$\boxed{\left|\int_a^b f(x)\,dx\right| \le \int_a^b |f(x)|\,dx}$$