Who's Up To the Challenge?80

$\begin{aligned}I_1&=\int_0^{\frac12}\cos(x-x^2)\{2019x\}dx+\int_{\frac12}^1\cos(x-x^2)\{2019x\}dx\\ &=\int_0^{\frac12}\cos(x-x^2)\{2019x\}dx-\int_{\frac12}^0\cos(1-u-(1-u)^2)\{2019(1-u)\}du&\text{by letting }u=1-x\\ &=\int_0^{\frac12}\cos(x-x^2)\{2019x\}dx+\int_0^{\frac12}\cos(u-u^2)\{2019(1-u)\}du\\ &=\int_0^{\frac12}\cos(x-x^2)\{2019x\}dx+\int_0^{\frac12}\cos(x-x^2)\{2019(1-x)\}dx&\text{since }u\text{ and }x\text{ are dummy variables}\\ &=\int_0^{\frac12}\cos(x-x^2)\left(\{2019x\}+\{2019(1-x)\}\right)dx\end{aligned}$

Now, the functions $\{2019x\}$ and $\{2019(1-x)\}$ can be rewritten as $2019x-n$ and $2019(1-x)-m$ where $n$ and $m$ are integers such that $0\leq2019x-n<1$ and $0\leq2019(1-x)-m<1$ .

$\begin{aligned}\{2019x\}+\{2019(1-x)&=2019x-n+2019(1-x)-m\\ &=2019x-n+2019-2019x-m\\ &=2019-n-m\end{aligned}$

We can see that $\{2019x\}+\{2019(1-x)\}$ must always be an integer, and since $0\leq{2019x}<1$ and $0\leq{2019(1-x)}<1$ this integer must be either 0 or 1. It can only be 0 if both $\{2019x\}$ and $\{2019(1-x)\}$ equal 0, which if possible would only occur in infinitesimal locations which are negligible to the integral, and so $\{2019x\}+\{2019(1-x)\}=1$ .

$\therefore I_1=\int_0^{\frac12}\cos(x-x^2)$

$\begin{aligned}I_2&=\int_0^{\frac12}\cos(x-x^2)dx+\int_{\frac12}^1\cos(x-x^2)dx\\ &=\int_0^{\frac12}\cos(x-x^2)dx-\int_{\frac12}^0\cos(1-u-(1-u)^2)du&\text{by letting }u=1-x\\ &=\int_0^{\frac12}\cos(x-x^2)dx+\int_0^{\frac12}\cos(u-u^2)du\\ &=\int_0^{\frac12}\cos(x-x^2)dx+\int_0^{\frac12}\cos(x-x^2)dx&\text{since }u\text{ and }x\text{ are dummy variables}\\ &=2\int_0^{\frac12}\cos(x-x^2)dx\\ &=2I_1\\ \therefore n&=\frac{2I_1}{I_1}=2\end{aligned}$ $\therefore I_3=\int_0^1\{2019x\}^2dx$ It is easy to see that the function $\{2019x\}$ equals $2019x$ for $0\leq x<\frac1{2019}$ , and then repeats every $\frac1{2019}$ units, repeating a total of 2019 times on the interval $[0,1]$ . Therefore we can write the integral as: $\begin{aligned}I_3&=2019\times\int_0^{\frac1{2019}}(2019x)^2dx\\ &=2019\times\left[2019^2\frac{x^3}3\right]_0^{\frac1{2019}}\\ &=2019\times2019^2\frac1{3\times2019^3}\\ &=\frac13\end{aligned}$ So the answer is $1+3=\boxed{4}$