Question

Find the Riemann sum for

f(x) = sin x
over the interval
[0, 2π],
where
x0 = 0, x1 = π/4, x2 = π/3, x3 = π, and x4 = 2π,
and where
c1 = π/6, c2 = π/3, c3 = 2π/3, and c4 = 3π/2.

Answer 1

I don't know for certain, but it looks like each
`c_i` is supposed to be the representative value of
`f(x)` over each subinterval
`[x_(i-1),x_i]`, so that the Riemann sum is given by


`\displaystyle\sum_(i=1)^4f(c_i)\Delta x_i`

where
`\Delta x_i=x_i-x_(i-1)`

The sum is then


`\displaystyle\sum_(i=1)^4f(c_i)\Delta x_i=\left(\frac\pi4-0\right)\sin\frac\pi6+\left(\frac\pi3-\frac\pi4\right)\sin\frac\pi3+\left(\pi-\frac\pi3\right)\sin\frac{2\pi}3+\left(2\pi-\pi\right)\sin\frac{3\pi}2`

`=\frac{(3\sqrt3-7)\pi}8\approx-0.7084`

Compare to the exact value of the corresponding definite integral,


`\displaystyle\int_0^(2\pi)\sin x=0`

Answer 2

The value of riemann sum is -0.70836.

Explanation:

The given function is

`f(x)=\sin x`

over the interval [0, 2π], where

`x_0=0,x_1=(\pi)/(4),x_2=(\pi)/(3),x_3=\pi,x_4=2\pi`

`c_1=(\pi)/(6),c_2=(\pi)/(3),c_3=(2\pi)/(3),c_4=(3\pi)/(2)`

Using Riemann sum formula

`\sum_(i=1)^(n)f(c_i)\Delta x_i,x_(i-1)\leq c_i\leq x_i`

where, n=4, so

`\sum_(i=1)^(4)f(c_i)\Delta x_i=f(c_1)\Delta x_1+f(c_2)\Delta x_2+f(c_3)\Delta x_3+f(c_4)\Delta x_4`

`f(c_1)(x_1-x_0)+f(c_2)(x_2-x_1)+f(c_3)(x_3-x_2)+f(c_4)(x_4-x_3)`

`f((\pi)/(6))((\pi)/(4)-0)+f((\pi)/(3))((\pi)/(3)-(\pi)/(4))+f((2\pi)/(3))(\pi-(\pi)/(3))+f((3\pi)/(2))(2\pi-\pi)`

`f((\pi)/(6))((\pi)/(4))+f((\pi)/(3))((\pi)/(12))+f((2\pi)/(3))((2\pi)/(3))+f((3\pi)/(2))(\pi)`

The given function is f(x)=sin x, so we get

`(1)/(2)((\pi)/(4))+(√(3))/(2)((\pi)/(12))+(√(3))/(2)((2\pi)/(3))+(-1)(\pi)`

`(\pi)/(8)+(√(3))/(24)\pi+(√(3))/(3)\pi-\pi`

`(3√(3))/(8)\pi+(7\pi)/(8)\approx -0.70836`

Therefore the value of riemann sum is -0.70836.