# Find the Riemann sum for f(x) = 3 sin x, 0 ≤ x ≤ 3π/2, with six terms, taking the sample points to be right endpoints.(a) Find the Riemann sum for f(x) = 3 sin x, 0 ≤ x ≤ 3π/2, with six...

Find the Riemann sum for f(x) = 3 sin x, 0 ≤ x ≤ 3*π*/2, with six terms, taking the sample points to be right endpoints.

(a) Find the Riemann sum for f(x) = 3 sin x, 0 ≤ x ≤ 3*π*/2, with six terms, taking the sample points to be right endpoints. (Round your answers to six decimal places.)

R6=____________

(b) Repeat part (a) with midpoints as the sample points.

M6=____________

### 1 Answer | Add Yours

You need to find the width of interval, hence you may use the formula:

`Delta x = (3pi/2 - 0)/6 =gt Delta x = 3pi/12 =gt Delta x = pi/4`

You need to find the 6 right endpoints such that:

x_1 = pi/4 ; x_2 = pi/4 + pi/4 = pi/2

`x_3 = pi/2 + pi/4 = 3pi/4`

`x_4 = 3pi/4 + pi/4 = pi`

`x_5 = pi+pi/4 = 5pi/4`

`x_6 = 5pi/4 + pi/4 = 6pi/4 =gt x_6 = 3pi/2`

You need to remember the formula of Riemann's sum such that:

`R_6 = sum_(i=1)^6 f(x_i)Delta x`

`R_6 = (pi/4)*(f(x_1)+f(x_2)+f(x_3)+f(x_4)+f(x_5)+f(x_6))`

`R_6 = (3pi/4)*(sin (pi/4)+sin (pi/2)+sin (3pi/4)+sin (pi)+sin (5pi/4)+sin (3pi/2))`

`R_6 = (3pi/4)*(sqrt2/2 + 1 + sqrt2/2 + 0 - sqrt2/2 - 1)`

`R_6 = (3pi/4)*(sqrt2/2) =gt R_6 = 1.666081`

**Hence, evaluating the Riemann's sum under given conditions yields `R_6 = 1.666081.` **

b)You need to find the midpoints of intervals `[0,pi/4],[pi/4,pi/2],[pi/2,3pi/4],[3pi/4,pi],[pi,5pi/4],[5pi/4,3pi/2] ` such that:

`m_1 = pi/8`

`m_2 = (pi/4+pi/2)/2 = 3pi/8`

`m_3 = (pi/2+3pi/4)/2 = 5pi/8`

`m_4 = (3pi/4+pi)/2 = 7pi/8`

`m_5 = 9pi/8`

`m_6 = 11pi/8`

You should evaluate Riemann's sum substituting `m_1,m_2,m_3,m_4,m_5,m_6` in equation of function such that:

`M_6 = (3pi/4)*(sin (pi/8)+sin (3pi/8)+sin (5pi/8)+sin (7pi/8)+sin (9pi/8)+sin (11pi/8))`

`M_6 = (3pi/4)*(0.382683+0.923879+0.923879+0.382683-0.382683-0.923879)`

`M_6 = 3.078514`

**Hence, evaluating the Riemann's sum under given conditions yields `M_6 = 3.078514.` **