Question

State Bonnet’s mean value theorem for integrals. Apply it to show that:

|₃∫⁵ cosxdx/x|≤ 2/3

Answer 1

Bonnet Mean Value Theorem .

Suppose $f$ is Lebesgue integrable on $[a, b]$ and $g:[a,b]\to\R$ is monotone. 

i) If $g$ is non-negative, decreasing and greater than or equal to $0,$ for $A\in \R,$ $A\geq \lim\limits_{x\to a^{+}}g(x)$ there exists $C$ such that $a\leq C\leq b$ and

$$\displaystyle\int_{a}^{b}f(x)g(x)dx=A\displaystyle\int_{a}^{C}f(x)dx$$

ii) If $g$ is non-negative, increasing and greater than or equal to $0,$ for $B\in \R,$ $A\geq \lim\limits_{x\to b^{-}}g(x)$ there exists $C$ such that $a\leq C\leq b$ and

$$\displaystyle\int_{a}^{b}f(x)g(x)dx=B\displaystyle\int_{C}^{b}f(x)dx$$

Consider

$$\displaystyle\int_{3}^{5}\dfrac{1}{x}\cos xdx$$

The function $f(x)=\cos x$ is integrable on $[3, 5].$

The function $g(x)=\dfrac{1}{x}$ is non-negative, monotone decreasing on $[3, 5].$

Then by the Bonnet Mean Value Theorem, for $A\geq \lim\limits_{x\to 3^{+}}\dfrac{1}{x}$ there exists $C$ such that $3\leq C\leq 5$ and

$$\displaystyle\int_{3}^{5}\dfrac{1}{x}\cos xdx=A\displaystyle\int_{3}^{C}\cos xdx$$

Let $A=\dfrac{1}{3}$

$$\bigg|\displaystyle\int_{3}^{5}\dfrac{1}{x}\cos xdx\bigg|=\dfrac{1}{3}\bigg|\displaystyle\int_{3}^{C}\cos xdx\bigg|$$

$$=\dfrac{1}{3}\big|[\sin x]\big|\begin{matrix} 5 \\ 3 \end{matrix}\leq\dfrac{1}{3}(2)=\dfrac{2}{3}$$

Therefore

$$\bigg|\displaystyle\int_{3}^{5}\dfrac{1}{x}\cos xdx\bigg|\leq\dfrac{2}{3}$$