Hot Integral - 31 (You missed these ?)

It can be rewritten as $\displaystyle \int\limits_{2018^2}^{2019^2}\left( x^{dx}-1\right) \frac{dx}{dx}$ $=\displaystyle \int\limits_{2018^2}^{2019^2}\left( \frac{x^{dx}-1}{dx}\right)dx$ $=\displaystyle \int\limits_{2018^2}^{2019^2}\left(\lim_{\Delta x \to 0} \frac{x^{\Delta x}-1}{\Delta x}\right)dx$ $=\displaystyle \int\limits_{2018^2}^{2019^2}(\ln x) dx$

Now you know ;)

$\begin{aligned}k&={\large\int}_{{2018}^2}^{{2019}^2}(x^{dx}-1)\\ &={\large\int}_{{2018}^2}^{{2019}^2}(e^{dxln(x)}-1)\\\\ &\text{as } dx\to0\\\\ e&=(1+dx)^{\tfrac{1}{dx}}\\\\ &\text{Substituting for e}\\\\ k&={\large\int}_{{2018}^2}^{{2019}^2}\left((1+dx)^{\tfrac{1}{dx}}\right)^{dxln(x)}-1\\ &={\large\int}_{{2018}^2}^{{2019}^2}(1+dx)^{ln(x)}-1\\\\ &\text{as } x\text{ varies from } { { 2018}^2}\text{ to }{{2019}^2}\\ &\ln(x) \text{ varies from } {15.2197244_\cdots} \text{ to } {15.2207152_\cdots}\\ &\text{ if } 0<t<<<<1 \text{ and } \text{ n is small}\\ &(1+t)^n\approx 1+nt \\ &\text{using the above approximation}\\ k&\approx{\large\int}_{{2018}^2}^{{2019}^2}(1+ln(x)dx)-1\\ &\approx{\large\int}_{{2018}^2}^{{2019}^2}ln(x)dx\\ &\approx x(ln(x)-1)\biggr|_{{2018}^2}^{{2019}^2}\\\\ k&\approx{61444.027740_\cdots}\\ \lfloor{k}\rfloor&=\color{#EC7300}\boxed{\color{#333333}61444}\end{aligned}$