L'hopital rule, but how?

By definition of floor function:

$x-1 \leq\lfloor x\rfloor \leq x~\forall x\in\mathfrak{R}$

$\implies \dfrac{1}{x}\leq\dfrac1{\lfloor x\rfloor}\leq \dfrac{1}{x-1}~\forall x\in\mathfrak R-\{0,1\}$

$\implies \dfrac{\ln x}{x}\leq\dfrac{\ln x}{\lfloor x\rfloor}\leq \dfrac{\ln x}{x-1}$

$(\small{\color{#D61F06}{\text{Since }\ln x>0\text{ as }x\to \infty}})$

$\implies \displaystyle\lim_{x\to\infty}\dfrac{\ln x}{x}\leq\displaystyle\lim_{x\to\infty}\dfrac{\ln x}{\lfloor x\rfloor}\leq\displaystyle\lim_{x\to\infty}\dfrac{\ln x}{x-1}$

By Squeeze theorem $\mathfrak{\color{#D61F06}{C}}$ is $0$ since:-

$\displaystyle\lim_{x\to\infty}\dfrac{\ln x}{x}=\displaystyle\lim_{x\to\infty}\dfrac{\ln x}{x-1}=0~~(\text{By }\color{#20A900}{L'hopital})$

And hence integration is simply:

$\Large\displaystyle\int_0^5(1)\mathrm{d}x~~~~~~~~~~(\small{\text{LOL}})$

$\huge =\color{#007fff}{\boxed{5}}~~~~~~~~~$

Let $\{x\}$ be the functional part of $x$ such that $x = \lfloor x \rfloor + \{x\}$ . This function has the property that $0 \leq \{x\} < 1$ and so $0 \leq \frac{\{x\}}{x} < \frac{1}{x}$ when $x>0$ . If we take the limit as $x$ goes to infinity then by the sandwich theorem, we have $\lim_{x \to \infty} \frac{\{x\}}{x} = 0$

$\begin{aligned} \lim_{x \to \infty} \frac{\ln x}{\lfloor x \rfloor} &= \lim_{x \to \infty} \frac{\frac{\ln x}{x}}{\frac{\lfloor x \rfloor}{x}} \\ &= \lim_{x \to \infty} \frac{\frac{\ln x}{x}}{\frac{x}{x} + \frac{\{x\}}{x}} \\ &= \lim_{x \to \infty} \frac{\frac{\ln x}{x}}{1 + 0} \\ &= \lim_{x \to \infty} \frac{\ln x}{x} \end{aligned}$

This is a standard limit which has value $0$ so therefore $\mathfrak{C} = 0$

After this, we simply have to calculate $\int_0^5 e^{\mathfrak{C} x} \,dx$ which equals $\int_0^5 {e^0x} \,dx = \int_0^5 dx = 5$

Write floor (x) as x-{x}, since x tends to infinity , x-{x} tends to x as {x} is very small. Thus $\frac {ln(x)}{floor(x)}=\frac{ln(x)}{x}$ , whose limit is 0 at infinity. Thus the required answer is integration of $e^{0}$ from 1 to 5, which is 5.

pls don't write next time to answer upto 2 decimals for integral answers. I have to recheck my calculation 5 times!!!!!!!. Nice question