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

$\displaystyle\sum_{i=1}^4f(c_i)\Delta x_i=\left(\dfrac\pi4-0\right)\sin\dfrac\pi6+\left(\dfrac\pi3-\dfrac\pi4\right)\sin\dfrac\pi3+\left(\pi-\dfrac\pi3\right)\sin\dfrac{2\pi}3+\left(2\pi-\pi\right)\sin\dfrac{3\pi}2$
$=\dfrac{(3\sqrt3-7)\pi}8\approx-0.7084$

Compare to the exact value of the corresponding definite integral,

$\displaystyle\int_0^{2\pi}\sin x=0$

Answer 2

Answer:

The value of riemann sum is -0.70836.

Step-by-step explanation:

The given function is

$f(x)=\sin x$

over the interval  [0, 2π],  where

$x_0=0,x_1=\frac{\pi}{4},x_2=\frac{\pi}{3},x_3=\pi,x_4=2\pi$

$c_1=\frac{\pi}{6},c_2=\frac{\pi}{3},c_3=\frac{2\pi}{3},c_4=\frac{3\pi}{2}$

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

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

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

$f(\frac{\pi}{6})(\frac{\pi}{4}-0)+f(\frac{\pi}{3})(\frac{\pi}{3}-\frac{\pi}{4})+f(\frac{2\pi}{3})(\pi-\frac{\pi}{3})+f(\frac{3\pi}{2})(2\pi-\pi)$

$f(\frac{\pi}{6})(\frac{\pi}{4})+f(\frac{\pi}{3})(\frac{\pi}{12})+f(\frac{2\pi}{3})(\frac{2\pi}{3})+f(\frac{3\pi}{2})(\pi)$

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

$\frac{1}{2}(\frac{\pi}{4})+\frac{\sqrt{3}}{2}(\frac{\pi}{12})+\frac{\sqrt{3}}{2}(\frac{2\pi}{3})+(-1)(\pi)$

$\frac{\pi}{8}+\frac{\sqrt{3}}{24}\pi+\frac{\sqrt{3}}{3}\pi-\pi$

$\frac{3\sqrt{3}}{8}\pi+\frac{7\pi}{8}\approx -0.70836$

Therefore the value of riemann sum is -0.70836.