Question

Limn→∞∑i=1nxicos(xi)Δx,[0,2π]

Answer 1

However the $x_i$ are chosen, the sum is equivalent to the definite integral,

$\displaystyle\lim_{n\to\infty}\sum_{i=1}^n x_i\cos x_i\Delta x=\int_0^{2\pi}x\cos x\,\mathrm dx$

which can be computed via integration by parts. If $u=x$ and $\mathrm dv=\cos x\,\mathrm dx$, you have

$\displaystyle\int_0^{2\pi}x\cos x\,\mathrm dx=[uv]_0^{2\pi}-\int_0^{2\pi}v\,\mathrm du$

or equivalently,

$\displaystyle\int_0^{2\pi}x\cos x\,\mathrm dx=[x\sin x]_0^{2\pi}-\int_0^{2\pi}\sin x\,\mathrm dx=[\cos x]_0^{2\pi}=0$