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`