Subsection 1.2.2 Optional — More properties of integration: inequalities for integrals
We are still unable to integrate many functions, however with a little work we can infer bounds on integrals from bounds on their integrands.
Theorem 1.2.13 Inequalities for Integrals
Let \(a\le b\) be real numbers and let the functions \(f(x)\) and \(g(x)\) be integrable on the interval \(a\le x\le b\text{.}\)
- If \(f(x)\ge 0\) for all \(a\le x\le b\text{,}\) then
\begin{gather*}
\int_a^b f(x)\,\dee{x} \ge 0
\end{gather*}
- If \(f(x)\le g(x)\) for all \(a\le x\le b\text{,}\) then
\begin{gather*}
\int_a^b f(x)\,\dee{x} \le \int_a^b g(x)\,\dee{x}
\end{gather*}
- If there are constants \(m\) and \(M\) such that \(m\le f(x)\le M\) for all \(a\le x\le b\text{,}\) then
\begin{gather*}
m(b-a)\le \int_a^b f(x)\,\dee{x} \le M(b-a)
\end{gather*}
- We have
\begin{gather*}
\bigg|\int_a^b f(x)\,\dee{x}\bigg|\le \int_a^b |f(x)|\,\dee{x}
\end{gather*}
Proof
- By interpreting the integral as the signed area, this statement simply says that if the curve \(y=f(x)\) lies above the \(x\)-axis and \(a\le b\text{,}\) then the signed area of \(\big\{\ (x,y)\ \big|\ a\le x\le b,\ 0\le y\le f(x)\ \big\}\) is at least zero. This is quite clear. Alternatively, we could argue more algebraically from Definition 1.1.9. We observe that when we define \(\int_a^b f(x)\dee{x}\) via Riemann sums, every summand, \(f(x_{i,n}^*)\,\frac{b-a}{n}\ge 0\text{.}\) Thus the whole sum is nonnegative and consequently, so is the limit, and thus so is the integral.
- We are assuming that \(g(x)-f(x)\geq 0\text{,}\) so part (a) gives
\begin{align*}
\int_a^b\big[g(x)-f(x)\big]\,\dee{x}\ge 0
\amp\implies \int_a^b g(x)\,\dee{x}-\int_a^b f(x)\,\dee{x}\ge 0 \\
\amp\implies \int_a^b f(x)\,\dee{x} \le \int_a^b g(x)\,\dee{x}
\end{align*}
- Applying part (b) with \(g(x)=M\) for all \(a\le x\le b\) gives
\begin{gather*}
\int_a^b f(x)\,\dee{x} \le \int_a^b M\,\dee{x} = M(b-a)
\end{gather*}
Similarly, viewing \(m\) as a (constant) function, and applying part (b) gives
\begin{gather*}
m\le f(x) \implies \overbrace{\int_a^bm\,\dee{x}}^{=m(b-a)} \le \int_a^b f(x)\,\dee{x}
\end{gather*}
- For any \(x\text{,}\) \(|f(x)|\) is either \(f(x)\) or \(-f(x)\) (depending on whether \(f(x)\) is positive or negative), so we certainly have
\begin{align*}
f(x)&\le |f(x)| & \text{and}&& -f(x)&\le |f(x)|
\end{align*}
Applying part (c) to each of those inequalities gives
\begin{align*}
\int_a^b f(x)\dee{x} &\le \int_a^b |f(x)|\dee{x} & \text{and} &&
-\int_a^b f(x)\dee{x} &\le \int_a^b |f(x)|\dee{x}
\end{align*}
Now \(\Big|\int_a^b f(x)\dee{x}\Big|\) is either equal to \(\int_a^b f(x)\dee{x}\) or \(-\int_a^b f(x)\dee{x}\) (depending on whether the integral is positive or negative). In either case we can apply the above two inequalities to get the same result, namely
\begin{align*}
\left|\int_a^b f(x)\dee{x}\right| &\leq \int_a^b |f(x)|\dee{x}.
\end{align*}
Example 1.2.14 \(\int_0^{\frac{\pi}{3}}\sqrt{\cos x}\dee{x}\)
Consider the integral
\begin{gather*}
\int_0^{\frac{\pi}{3}}\sqrt{\cos x}\dee{x}
\end{gather*}
This is not so easy to compute exactly 3 , but we can bound it quite quickly.
For \(x\) between \(0\) and \(\frac{\pi}{3}\text{,}\) the function \(\cos x\) takes values 4 between \(1\) and \(\frac{1}{2}\text{.}\) Thus the function \(\sqrt{\cos x}\) takes values between \(1\) and \(\frac{1}{\sqrt{2}}\text{.}\) That is
\begin{align*}
\frac{1}{\sqrt{2}} &\le \sqrt{\cos x} \le 1 & \text{for $0\le x\le \frac{\pi}{3}$}.
\end{align*}
Consequently, by Theorem 1.2.13(b) with \(a=0\text{,}\) \(b=\frac{\pi}{3}\text{,}\) \(m= \frac{1}{\sqrt{2}}\) and \(M=1\text{,}\)
\begin{align*}
\frac{\pi}{3\sqrt{2}} &\le \int_0^{\frac{\pi}{3}} \sqrt{\cos x}\dee{x} \le \frac{\pi}{3}\\
\\
\end{align*}
Plugging these expressions into a calculator gives us
\begin{align*}
0.7404804898 & \le \int_0^{\frac{\pi}{3}} \sqrt{\cos x}\dee{x} \leq 1.047197551
\end{align*}