Chapter 6.3, Problem 38E

To determine

The Max-Min inequality for lower and upper bound values.

Answer to Problem 38E

The value of the integral is $\frac{49}{68}\le {\displaystyle \underset{0}{\overset{1}{\int }}\frac{1}{1+x^4}dx}\le \frac{33}{34}$ .

Explanation of Solution

Given information:

  ${\displaystyle \underset{0}{\overset{0.5}{\int }}\frac{1}{1+x^4}dx\text{and}{\displaystyle \underset{0.5}{\overset{1}{\int }}\frac{1}{1+x^4}}}dx$

Formula used:

  $\min f\left(b-a\right)\le {\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx\le \max f\left(b-a\right)$

Calculation:

If max f and min fare the maximum and minimum values of f on [a, b], then

  $\min f\left(b-a\right)\le {\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx\le \max f\left(b-a\right)$

Here,

  $f\left(x\right)=\frac{1}{1+x^4}$

  

Observe the graph in the interval [0, 0.5], function is decreasing.

So,

  $\begin{array}{l}\min f=f\left(0.5\right)\\ =\frac{1}{1+{\left(\frac{1}{2}\right)}^4}\\ =\frac{16}{17}\\ \max f=f\left(0\right)\\ =\frac{1}{1+0^4}\\ \max f=1\end{array}$

So,

  $\begin{array}{l}\frac{16}{17}\left(0.5-8\right)\le {\displaystyle \underset{0}{\overset{0.5}{\int }}\frac{1}{1+x^4}dx\le 1\left(0.5-0\right)}\\ \frac{8}{17}\le {\displaystyle \underset{0}{\overset{0.5}{\int }}\frac{1}{1+x^4}dx\le 0.5}\end{array}$

If max f and min fare the maximum and minimum values of f on [a, b], then

  $\min f\left(b-a\right)\le {\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx\le \max f\left(b-a\right)$

Here,

  $f\left(x\right)=\frac{1}{1+x^4}$

Observe the graph in the interval [0.5, l], function is decreasing.

So,

  $\begin{array}{l}\min f=f\left(1\right)\\ =\frac{1}{1+1^4}\\ \min f=\frac{1}{2}\\ \max f=f\left(0.5\right)\\ =\frac{1}{1+{\left(\frac{1}{2}\right)}^4}\\ \max f=\frac{16}{17}\end{array}$

So,

  $\begin{array}{l}\frac{1}{2}\left(1-0.5\right)\le {\displaystyle \underset{0.5}{\overset{1}{\int }}\frac{1}{1+x^4}dx\le \frac{16}{17}\left(1-0.5\right)}\\ \frac{1}{4}\le {\displaystyle \underset{0.5}{\overset{1}{\int }}\frac{1}{1+x^4}dx\le \frac{8}{17}}\end{array}$

Now, add both obtained above inequalities.

  $\frac{8}{17}+\frac{1}{4}\le {\displaystyle \underset{0}{\overset{0.5}{\int }}\frac{1}{1+x^4}dx+{\displaystyle \underset{0.5}{\overset{1}{\int }}\frac{1}{1+x^4}dx\le \frac{1}{2}+\frac{8}{17}}}$

Now, apply this rule of definite integral

  ${\displaystyle \underset{a}{\overset{c}{\int }}f\left(x\right)}dx={\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx+{\displaystyle \underset{b}{\overset{c}{\int }}f\left(x\right)dx}$

So,

  $\begin{array}{l}\frac{32+17}{68}\le {\displaystyle \underset{0}{\overset{1}{\int }}\frac{1}{1+x^4}dx\le \frac{17+16}{34}}\\ \frac{49}{68}\le {\displaystyle \underset{0}{\overset{1}{\int }}\frac{1}{1+x^4}dx}\le \frac{33}{34}\end{array}$

Conclusion:

The value of the integral is $\frac{49}{68}\le {\displaystyle \underset{0}{\overset{1}{\int }}\frac{1}{1+x^4}dx}\le \frac{33}{34}$ .