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.
Your answer sucks, Steve :/
To find the Riemann sum for a function over a given interval, you can use the formula:
Rn = Σf(ci) Δxi
where:
- Rn is the Riemann sum using n subintervals
- f(ci) is the value of the function at the point ci within each subinterval
- Δxi is the width of each subinterval
Let's calculate the Riemann sum for the function f(x) = sin(x) over the interval [0, 2π] using the provided values of xi and ci.
First, let's calculate Δxi for each subinterval.
- Δx1 = x1 - x0 = π/4 - 0 = π/4
- Δx2 = x2 - x1 = π/3 - π/4 = π/12
- Δx3 = x3 - x2 = π - π/3 = 2π/3 - π/3 = π/3
- Δx4 = x4 - x3 = 2π - π = π
Now, let's substitute the values into the Riemann sum formula:
Rn = f(c1) Δx1 + f(c2) Δx2 + f(c3) Δx3 + f(c4) Δx4
Substituting the values, we have:
Rn = sin(π/6) (π/4) + sin(π/3) (π/12) + sin(2π/3) (π/3) + sin(3π/2) (π)
Now, we can calculate each term:
sin(π/6) = 1/2
sin(π/3) = √3/2
sin(2π/3) = √3/2
sin(3π/2) = -1
Substituting these values and simplifying, we have:
Rn = (1/2)(π/4) + (√3/2)(π/12) + (√3/2)(π/3) - (π)
Simplifying further:
Rn = π/8 + (√3/24)π + (√3/6)π - π
Combining like terms:
Rn = (1/8 + √3/24 + √3/6 - 1)π
To simplify further, we need to find a common denominator:
Rn = [(3 + √3 + 8√3 - 24)/24]π
Simplifying the numerator:
Rn = [(11 + 9√3)/24]π
Therefore, the Riemann sum for f(x) = sin(x) over the interval [0, 2π] using the given values is:
Rn = [(11 + 9√3)/24]π
there are many Riemann Sum calculators online.
google is your friend