Calculus : Integral Inequality?

Calculus is one of the major topics in Mathematics across the world. . The problems involving the definite integrals require advanced knowledge. In this lecture with the help of the problem defined here we try to get the insights of the definite integrals calculations.

Here we solve a problem involving integral inequality as if $a<\underset{\mathit{\pi }/3}{\overset{\mathit{\pi }/8}{\int }}\frac{x}{\cos x}\mathrm{dx}<b$ and $a,b\in \frac{1}{2}\mathbb{Z}$. Find $a^2+b^2$.

It is a very distinct unique tricky problem. We solve it without practically finding the value of integral. Kindly go through below paragraphs for full solution. This type of problems of importance will be solved regularly by IITian here.

What is meant by Integral Inequality in Calculus ?

Consider an integral inequality like,

$a<f(x)<b\Rightarrow a(q-p)<\underset{p}{\overset{q}{\int }}f(x)\mathrm{dx}<b(q-p)$

To understand the above inequality, let us draw an analogy to simple algebraic inequality such as ,

(x-1) < x < (x+1) for any real value of x.

The proof of above simple inequality is ,

First consider left part of the inequality, (x-1)<x implies , -1< 0. This is always true.

Similarly if the right part of the inequality is taken, x<(x+1) implies , 0<1. This is always true.

Let us expand our this understanding to integral inequality we just considered. From the basic principles of definite integrals,

$a<f(x)<b\Rightarrow a(q-p)<\underset{p}{\overset{q}{\int }}f(x)\mathrm{dx}<b(q-p)$ can also be proved very easily.

Problem on Integral Inequality:

Let us take a problem in understanding the concepts of integral inequality and solve it.

let $a<\underset{\mathit{\pi }/3}{\overset{\mathit{\pi }/8}{\int }}\frac{x}{\cos x}\mathrm{dx}<b$ and $a,b\in \frac{1}{2}\mathbb{Z}$. Find $a^2+b^2$.

Solution

The equation in the question is given in the below image.

Let us solve the problem step by step.

The lower limit is greater than upper limit as $\mathit{\pi }/3>\mathit{\pi }/8$ .

Hence, $\underset{\mathit{\pi }/3}{\overset{\mathit{\pi }/8}{\int }}\frac{x}{\cos x}\mathrm{dx}=-\underset{\mathit{\pi }/8}{\overset{\mathit{\pi }/3}{\int }}x\sec \mathrm{xdx}$ for the given limits.

From above limits, $x>0,\cos x>0$ and $\underset{\mathit{\pi }/8}{\overset{\mathit{\pi }/3}{\int }}x\sec \mathrm{xdx}>0$.

From above, $\underset{\mathit{\pi }/3}{\overset{\mathit{\pi }/8}{\int }}\frac{x}{\cos x}\mathrm{dx}<0$.

For the given interval, $\cos x\ge \cos \mathit{\pi }/3$=1/2.

Hence $\sec x\le 2$ and $\underset{\mathit{\pi }/8}{\overset{\mathit{\pi }/3}{\int }}x\sec \mathrm{xdx}<2\underset{\mathit{\pi }/8}{\overset{\mathit{\pi }/3}{\int }}\mathrm{xdx}$.

From above,

$\underset{\mathit{\pi }/8}{\overset{\mathit{\pi }/3}{\int }}x\sec \mathrm{xdx}<{{\displaystyle {[x}^2]}}_{\mathit{\pi }/8}^{\mathit{\pi }/3}=\frac{{\mathit{\pi }}^2}{9}-\frac{{\mathit{\pi }}^2}{64}=\frac{55{\mathit{\pi }}^2}{576}$ <1

Hence , $-1<\underset{\mathit{\pi }/3}{\overset{\mathit{\pi }/8}{\int }}\frac{x}{\cos x}\mathrm{dx}<0$

Above when related to the problem equation, $a<\underset{\mathit{\pi }/3}{\overset{\mathit{\pi }/8}{\int }}\frac{x}{\cos x}\mathrm{dx}<b$ ,

a=-1 and b=0.

Hence $a^2+b^2=(-1{)}^2+0^2=1$ Answer.

The above is the solution if a and b are assunmed to be integers. If the rational numbers re allowed, the solution can be ,

a=-1, and b= -1/2.

From above if rational numbers allowed, the answer will be 5/4.

The concepts of rational numbers is explained in video from my channel maths& science by iitian.

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