# Prove the following?

## Prove ${\int}_{1}^{2} \left(\frac{{e}^{x} - \ln x}{x} ^ 2 - 1\right) \mathrm{dx} > 0$

May 22, 2018

Check below.

#### Explanation:

${\int}_{1}^{2} \left(\frac{{e}^{x} - \ln x}{x} ^ 2 - 1\right) \mathrm{dx} > 0$ $\iff$

${\int}_{1}^{2} \left(\frac{{e}^{x} - \ln x}{x} ^ 2\right) \mathrm{dx} > {\int}_{1}^{2} \left(1\right) \mathrm{dx}$ $\iff$

${\int}_{1}^{2} \left(\frac{{e}^{x} - \ln x}{x} ^ 2\right) \mathrm{dx} > {\left[x\right]}_{1}^{2}$ $\iff$ $\iff$

${\int}_{1}^{2} \left(\frac{{e}^{x} - \ln x}{x} ^ 2\right) \mathrm{dx} > 2 - 1$ $\iff$

${\int}_{1}^{2} \left(\frac{{e}^{x} - \ln x}{x} ^ 2\right) \mathrm{dx} > 1$

We need to prove that

${\int}_{1}^{2} \left(\frac{{e}^{x} - \ln x}{x} ^ 2\right) \mathrm{dx} > 1$

Consider a function $f \left(x\right) = {e}^{x} - \ln x$ , $x > 0$

From the graph of ${C}_{f}$ we can notice that for $x > 0$

we have ${e}^{x} - \ln x > 2$

Explanation:

$f \left(x\right) = {e}^{x} - \ln x$ , $x$$\in$$\left[\frac{1}{2} , 1\right]$

$f ' \left(x\right) = {e}^{x} - \frac{1}{x}$

$f ' \left(\frac{1}{2}\right) = \sqrt{e} - 2 < 0$

$f ' \left(1\right) = e - 1 > 0$

According to Bolzano (Intermediate Value) Theorem we have $f ' \left({x}_{0}\right) = 0$ $\iff$ ${e}^{{x}_{0}} - \frac{1}{x} _ 0 = 0$ $\iff$

${e}^{{x}_{0}} = \frac{1}{x} _ 0$ $\iff$ ${x}_{0} = - \ln {x}_{0}$

The vertical distance is between ${e}^{x}$ and $\ln x$ is minimum when $f \left({x}_{0}\right) = {e}^{{x}_{0}} - \ln {x}_{0} = {x}_{0} + \frac{1}{x} _ 0$

We need to show that $f \left(x\right) > 2$ , $\forall x$$> 0$

$f \left(x\right) > 2$ $\iff$ ${x}_{0} + \frac{1}{x} _ 0 > 2$ $\iff$

${x}_{0}^{2} - 2 {x}_{0} + 1 > 0$ $\iff$ ${\left({x}_{0} - 1\right)}^{2} > 0$ $\to$ true for $x > 0$

graph{e^x-lnx [-6.96, 7.09, -1.6, 5.42]}

$\frac{{e}^{x} - \ln x}{x} ^ 2 > \frac{2}{x} ^ 2$

${\int}_{1}^{2} \left(\frac{{e}^{x} - \ln x}{x} ^ 2\right) \mathrm{dx} > {\int}_{1}^{2} \left(\frac{2}{x} ^ 2\right) \mathrm{dx}$ $\iff$

${\int}_{1}^{2} \left(\frac{{e}^{x} - \ln x}{x} ^ 2\right) \mathrm{dx} > {\left[- \frac{2}{x}\right]}_{1}^{2}$ $\iff$

${\int}_{1}^{2} \left(\frac{{e}^{x} - \ln x}{x} ^ 2\right) \mathrm{dx} >$$- 1 + 2$ $\iff$

${\int}_{1}^{2} \left(\frac{{e}^{x} - \ln x}{x} ^ 2\right) \mathrm{dx} > 1$ $\iff$

${\int}_{1}^{2} \left(\frac{{e}^{x} - \ln x}{x} ^ 2 - 1\right) \mathrm{dx} > 0$