Chapter 5.3, Problem 71E

a.

To determine

To show: $1\le \sqrt{1+x^3}\le 1+x^3$ for $x\ge 0$

Explanation of Solution

Given information : $x\ge 0$

Concept used : $f\left(x\right)$ is said to be increasing or decreasing function according to $f^{\prime }\left(x\right)>0$ or $f^{\prime }\left(x\right)<0$ respectively.

If $f\left(x\right)$ is increasing function then $a>b\Rightarrow f\left(a\right)>f\left(b\right)$

Proof:

  $f\left(x\right)=\sqrt{1+x^3}-\left(1+x^3\right)$ (say)

Now,

  $\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{3x^2}{2\sqrt{1+x^3}}-3x^2\\ =x^2\left[\frac{3}{2\sqrt{1+x^3}}-3\right]\end{array}$

When $x$ is increasing, then $\frac{1}{\sqrt{1+x^3}}$ is decreasing.

So, $f^{\prime }\left(x\right)<0\forall x\ge 0$

Hence, $f\left(x\right)$ is decreasing function.

Therefore, for $x\ge 0$

  $f\left(x\right)\le 0$

  $\begin{array}{l}\Rightarrow \sqrt{1+x^3}-\left(1+x^3\right)\le 1-1\\ \Rightarrow \sqrt{1+x^3}\le \left(1+x^3\right)\mathrm{.........}\left(i\right)\end{array}$

Again, $g\left(x\right)=\sqrt{1+x^3}-1$ (say)

Then, $g^{\prime }\left(x\right)=\frac{3x^2}{2\sqrt{1+x^3}}>0\forall \text{ x>0}$

  $\Rightarrow \text{g}\left(x\right)$ is increasing function.

So, $x\ge 0$

  $\begin{array}{l}\Rightarrow g\left(x\right)\ge g\left(0\right)\\ \Rightarrow \sqrt{1+x^3}-1\ge 1-1\\ \Rightarrow \sqrt{1+x^3}\ge 1\mathrm{................}\left(ii\right)\end{array}$

Combining (i) and (ii) we get,

  $1\le \sqrt{1+x^3}\le 1+x^3\left(Shown\right)$

b.

To determine

To show: $1\le {\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{1+x^3}dx}\le 1.25$

Explanation of Solution

Given information : ${\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{1+x^3}dx}$

Concept used :

If $f_1\left(x\right)<f\left(x\right)<f_2\left(x\right)$ for $x\in \left[a,b\right]$

Then, ${\displaystyle \underset{a}{\overset{b}{\int }}{f}_1\left(x\right)}<{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}<{\displaystyle \underset{a}{\overset{b}{\int }}{f}_2\left(x\right)}$

Proof:

From part (a) we have,

  $1\le \sqrt{1+x^3}\le 1+x^3$ for $x\in \left[a,b\right]$

So,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{1}{\int }}1dx}\le {\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{1+x^3}dx}\le {\displaystyle \underset{0}{\overset{1}{\int }}\left(1+x^3\right)dx}\\ \Rightarrow {\left[x\right]}_0^1\le {\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{1+x^3}dx}\le {\left[x\right]}_0^1+{\left[\frac{x^4}{4}\right]}_0^1\\ \Rightarrow 1-0\le {\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{1+x^3}dx}\le 1+\frac{1}{4}\\ \Rightarrow 1\le {\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{1+x^3}dx}\le 1.25\left(\text{Proved}\right)\end{array}$