Integral of the Year

First, notice that $\displaystyle\int_0^{2014} f(x) \,\mathrm{d}x$ is a constant, which we will call $C$ .

So we can write the given equation as $f(x)= 2x + C$ .

Now, if we integrate both sides of this equation from 0 to 2014 (w/respect to x), then we will get $C$ on the left side and eliminate $f(x)$ from the equation, allowing us to find $C$ .

$\begin{aligned} \displaystyle\int_0^{2014}f(x) \,\mathrm{d}x & = \int_0^{2014} 2x\,\mathrm{d}x + \int_0^{2014} C \,\mathrm{d}x \\ C &=2014^2 +2014C \end{aligned}$

So $\,\displaystyle C=\frac{-2014^2}{2013}$ , and $\,\displaystyle f(x) = 2x-\dfrac{2014^2}{2013}$ .

Finally, $\displaystyle \int_0^{2013} f(x) \,\mathrm{d}x = \int_0^{2013} \left(2x-\dfrac{2014^2}{2013}\right) \,\mathrm{d}x = 2013^2 - 2014^2 = \boxed{-4027}$ .