# Find the Riemann sum for f(x)=7sinx, x is between 0 and 3pi/2, with 6 intervalsFind the Riemann sum for f(x)=7sinx, 0 ≤ x ≤ 3π/2, with six terms, taking the sample points to be right endpoints...

Find the Riemann sum for f(x)=7sinx, x is between 0 and 3pi/2, with 6 intervals

Find the Riemann sum for f(x)=7sinx, 0 ≤ x ≤ 3*π*/2,

with six terms, taking the sample points to be right endpoints and midpoints. (Round your answers to six decimal places.)

### 1 Answer | Add Yours

You need to evaluate the following Riemann sum, using 6 subintervals of equal lengths, such that:

`S_6 = f(x_1)*Delta x + f(x_2)*Delta x + f(x_3)*Delta x + f(x_4)*Delta x + f(x_5)*Delta x + f(x_6)*Delta x`

`x_1,x_2,x_3,x_4,x_5,x_6` represent the right endpoint of each of the 6 subintervals.

Delta x represents the length of each of the 6 subintervals

You may evaluate the length of subinterval `Delta x` , such that:

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

You need to evaluate the right endpoints, such that:

`x_1 = 0 + pi/4 = 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 = (3pi)/2`

You need to evaluate `f(x_1),f(x_2),f(x_3),f(x_4),f(x_5),f(x_6)` such that:

`f(pi/4) = 7sin (pi/4) => f(pi/4) = 7sqrt2/2`

`f(pi/2) = 7sin (pi/2) => f(pi/2) = 7`

`f((3pi)/4) = 7sin ((3pi)/4) => f((3pi)/4) = 7sqrt2/2`

`f(pi) = 7sin pi => f(pi) = 0`

`f((5pi)/4) = 7sin ((5pi)/4) =>f((5pi)/4) = -7sqrt2/2`

`f((3pi)/2) = 7sin ((3pi)/2) => f((3pi)/2) = -7`

`S_6 = (7sqrt2/2)(pi/4) + (7)(pi/4) + (7sqrt2/2)(pi/4) + 0 + (-7sqrt2/2)(pi/4) + (-7)(pi/4)`

Reducing duplicate terms yields:

`S_6 = (7sqrt2 pi)/8`

**Hence, evaluating the Riemann sum, using the partition in 6 subintervals of equal lengths, yields **`S_6 = (7sqrt2 pi)/8.`