3.5 Comparison Properties of the Definite Integral

3.5.1 Basic Properties

 
Proposition 1. Let \(f\) and \(g\) be integrable functions on the closed interval \([a,b]\).

(iii) Assume that \(m\) and \(m\) are constants such that \(m \leq f(x) \leq M \) for all \(x\in [a,b]\). Then
\[\displaystyle m(b-a)\leq \int_a^b f(x) \,dx \leq M(b-a).\]

Proof. (i) Let \(\displaystyle \Delta x = \frac{b-a}{n}\) and \(x_i=a+i\Delta x\), be a partition of the interval \([a,b]\). Since \(f(x) \geq 0\), it follows that \(f(x_i)\Delta x \geq 0\), for \(i = 1,2, \dots , n\). Thus
\[ \sum_{i=1}^{n} f(x_i)\Delta x \geq 0 \;\; \Longrightarrow \;\;\int_a^bf(x)\,dx = \lim_{n\to \infty} \sum_{i=1}^{n} f(x_i)\Delta x \geq 0.\]

(ii) Set \(h(x) = f(x)-g(x)\). Then \(h(x) \geq 0\), and, by (i), \(\displaystyle \int_a^b h(x)\,dx \geq 0\), that is,
\[ \int_a^b (f(x)-g(x))\,dx \geq 0 \underset{(2)}{\Longrightarrow} \int_a^b f(x)\,dx – \int_a^bg(x)\,dx \geq 0 \Longrightarrow \int_a^b f(x)\,dx \geq \int_a^bg(x)\,dx.\]

(iii) This follows immediately from (ii) (by applying (ii) twice).

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3.5.2 Applications of Comparison Properties

Example 1. Show that \(\displaystyle 2 \leq \int_0^1 \sqrt{4+\sin^2 x} \,dx \leq \sqrt 5 \).

Solution. Given that \(\displaystyle 0 \leq \sin^2 x \leq 1 \), then
\[4 \leq 4+\sin^2 x < 4+1 \underset{?}{\Longrightarrow} 0 \leq \sqrt{4+\sin^2x} \leq \sqrt{5} \Longrightarrow\] \[\underset{\text{(iii)}}{\Longrightarrow} \int_0^1 2 \,dx \leq \int_0^1 \sqrt{4+\sin^2 x} \,dx \leq \int_0^1 \sqrt{5}\,dx.\] Example 2. Show that \(\displaystyle 1 \leq \int_0^1 \sqrt{1+x^6} \,dx \leq \sqrt{2}\).

Solution. Let \( x\in (0,1)\), then
\[\displaystyle 1 < 1+x^6 < 2 \Longrightarrow 1 < \sqrt{1+x^6} < \sqrt{2} \; \Longrightarrow\int_0^1 1\,dx \leq\int_0^1\sqrt{1+x^6}\,dx \leq\int_0^1\sqrt{2}\,dx ,\] and the result follows.  

Example 3. Show that \(\displaystyle -\frac{1}{5} \leq \int_1^2 \frac{1}{x^2-9}\,dx \leq -\frac{1}{8}\).

Solution.
\[\text{If } \; x\in[1,2] \underset{?}{\Longrightarrow } -8\leq x^2-9 \leq -5 \Longrightarrow -\frac{1}{5}\leq \frac{1}{x^2-9} \leq -\frac{1}{8}\Longrightarrow\]
\[ \Longrightarrow \int_1^2-\frac{1}{5}\,dx \leq \int_1^2 \frac{1}{x^2-9}\,dx \leq \int_1^2 -\frac{1}{8}\,dx \underset{?}{\Longrightarrow } -\frac{1}{5} \leq \int_1^2 \frac{1}{x^2-9}\,dx \leq -\frac{1}{8}.\]

In the following example, we use the fact that \(y=\cos x\) is a decreasing function on \(\left [0, \pi\right ]\). To see this, use the unit circle, or, sketch its graph.

Example 4. Show that \(\displaystyle
\frac{1}{6\sqrt{2}}\left ( -3+2\sqrt 3\right )\sqrt \pi
\leq
\int_{\sqrt{\pi/4}}^{\sqrt{\pi/3}}\cos t^2\,dt
\leq
\frac{1}{12}\left ( -3+2\sqrt 3\right )\sqrt \pi\).

Solution.
\[\text{If } \; \sqrt {\frac{\pi}{4}} \leq t \leq \sqrt{\frac{ \pi}{ 3}} \Longrightarrow
\frac{\pi}{4} \leq t^2 \leq \frac{\pi}{3} \underset{?}{\Longrightarrow}
\cos\frac{\pi}{3} \geq \cos t^2 \geq \cos \frac{\pi}{4} \Longrightarrow \frac{1}{\sqrt 2}\leq \cos t^2 \leq \frac{1}{2}.\]
The result follows directly from the last inequality.

Example 5. (a) Find values \(m\) and \(M\) such that \(m \leq \arcsin x \leq M\) for all \(\displaystyle x\in \left [ -\frac{1}{2}, \frac{1}{\sqrt 2}\right ]\);

(b) estimate \(\displaystyle \int_{-1/2}^{1/\sqrt 2}\arcsin x\,dx\).

Solution. Since \(y=\arcsin x\) is an increasing function (why),
\[\text{If } \, -\frac{1}{2} \leq x \leq \frac{1}{\sqrt 2}\Longrightarrow
\arcsin \left (-\frac{1}{2} \right ) \leq \arcsin x \leq \arcsin \frac{1}{\sqrt 2} \Longrightarrow\]
\[\underset{?}{\Longrightarrow}\int_{-1/2}^{1/\sqrt 2}-\frac{\pi}{6} \,dx \leq \int_{-1/2}^{1/\sqrt 2}\arcsin x\,dx \leq
\int_{-1/2}^{1/\sqrt 2}\frac{\pi}{4} \,dx \Longrightarrow\]
\[\underset{?}{\Longrightarrow} -\frac{1}{12}\left (1+\sqrt 2\right )\pi \leq \int_{-1/2}^{1/\sqrt 2}\arcsin x\,dx \leq \frac{1}{8}\left (1+\sqrt 2\right )\pi.\]