Chapter 6.3, Problem 37E

To determine

To Find: Upper and lower bounds for the given value.

Answer to Problem 37E

The upper and lower bounds of the value are $1$ and $\frac{1}{2}$ respectively.

Explanation of Solution

Given information :

  ${{\displaystyle \int }}_0^1\frac{1}{1+x^4}dx$

Calculation :

If max f and min f are the maximum and minimum values of f on $\left[a,b\right]$ , then

  $\min f\cup (b-a)\le {{\displaystyle \int }}_a^{b}f(x)dx\le \max f(b-a)$

Here,

  $f(x)=\frac{1}{1+x^4}$

  

Observe the graph in the interval $\left[0,1\right]$ , function is decreasing.

  $\begin{array}{cc}\text{SO},& \\ \min f& =f(1)\\ & =\frac{1}{1+1^4}\\ & =\frac{1}{2}\\ \max f& =f(0)\\ & =\frac{1}{1+0^4}\\ & =1\end{array}$

So,

  $\begin{array}{c}\frac{1}{2}(1-0)\le {{\displaystyle \int }}_0^1\frac{1}{1+x^4}dx\le 1(1-0)\\ \frac{1}{2}\le {{\displaystyle \int }}_0^1\frac{1}{1+x^4}dx\le 1\end{array}$