Fun with 2015

Let $g(y) = f(x)$ , where $y = x-2013$ . Then we have:

$\begin{cases} f(2013) = g(0) = 3 \\ f(2015) = g(2) = 1 \\ f(2017) = g(4) = 5 \\ f(2019) = g(6) = 2015 \end{cases} \text{ and } \displaystyle \int_{2013}^{2017} {f(x) dx} = \int_{0}^{4} {g(y) dy}$

Let $g(y) = ay^3+by^2+cy+d$ , then we have:

$\begin{cases} g(0) = (0)a +(0)b+(0)c+d = 3 \\ g(2) = 8a +4b+2c+d = 1 \\ g(4) = 64a +16b+4c+d = 5 \\ g(6) = 216a +36b+6c+d = 2015 \end{cases}$

Solving the equations, we get: $\space a = \frac{125}{3}$ , $b = -\frac{997}{4}$ , $c = \frac{1985}{6}$ and $d = 3$ .

$\begin{aligned} \Rightarrow \int_{2013}^{2017} f(x) dx & = \int_{0}^{4} g(y) dy \\ & = \int_{0}^{4} {\left(\frac{125}{3}y^3 -\frac{997}{4} y^2 + \frac{1985}{6}y + 3 \right) dy} \\ & = \left[\frac{125}{12}y^4 -\frac{997}{12} y^3 + \frac{1985}{12}y^2 + 3y \right]_0^4 \\ & = \boxed{8} \end{aligned}$

Simpson's Rule gives the exact integral of cubic function on any interval. So we calculate $(b-a)/6 *(f(2013)+4f(2015)+f(2017))=8$

Using finite differences , an exact cubic polynomial fitting those data points was computed. The resulting resulting polynomial was integrated as specified in the problem. $\text{generator}=\text{Function}\left[\{\text{diffs},\text{start},\text{step}\}, \\ \text{Function}\left[x,\text{Evaluate}\left[\text{Expand}\left[\sum _{i=0}^{\text{Length}[\text{diffs}]-1} \frac{\text{diffs}[[i+1]] \prod _{j=0}^{i-1} (n-j)}{i!}\text{/.}\, n\to \frac{x-\text{start}}{\text{step}}\right]\right]\right]\right]$ $\text{estimate}=\text{generator}(\{3,-2,6,2000\},2013,2) \Longrightarrow \\ \text{Function}\left[x,\frac{125 x^3}{3}-\frac{1007497 x^2}{4}+\frac{1522574809 x}{3}-\frac{1363545375851}{4}\right]$ $\int_{2013}^{2017} \text{estimate}(x) \, dx \Longrightarrow 8$

The computer algebra system used was Wolfram Mathematica ${}^{\text{TM}}$ 11.3.0.