Math Problem Statement

If f(x) = 3 cos(x) 0 ≤ x ≤ 3𝜋/4 evaluate the Riemann sum with n = 6, taking the sample points to be left endpoints. (Round your answer to six decimal places.)

Solution

To evaluate the Riemann sum for the function f(x)=3cos(x)f(x) = 3 \cos(x) over the interval 0x3π40 \leq x \leq \frac{3\pi}{4} using n=6n = 6 subintervals and left endpoints as sample points, we can follow these steps:

Step 1: Find the width of each subinterval.

The total interval is [0,3π4]\left[ 0, \frac{3\pi}{4} \right]. To find the width of each subinterval, we divide the total length by nn:

Δx=3π406=3π24=π8\Delta x = \frac{\frac{3\pi}{4} - 0}{6} = \frac{3\pi}{24} = \frac{\pi}{8}

Step 2: Identify the left endpoints of each subinterval.

The left endpoints of the subintervals will be:

x0=0,x1=π8,x2=2π8=π4,x3=3π8,x4=4π8=π2,x5=5π8x_0 = 0, \quad x_1 = \frac{\pi}{8}, \quad x_2 = \frac{2\pi}{8} = \frac{\pi}{4}, \quad x_3 = \frac{3\pi}{8}, \quad x_4 = \frac{4\pi}{8} = \frac{\pi}{2}, \quad x_5 = \frac{5\pi}{8}

Step 3: Evaluate the function f(x)=3cos(x)f(x) = 3 \cos(x) at each left endpoint.

Now we evaluate f(x)f(x) at each left endpoint:

f(x0)=3cos(0)=3×1=3f(x_0) = 3 \cos(0) = 3 \times 1 = 3 f(x1)=3cos(π8)f(x_1) = 3 \cos\left(\frac{\pi}{8}\right) f(x2)=3cos(π4)=3×223×0.70710678118=2.121320f(x_2) = 3 \cos\left(\frac{\pi}{4}\right) = 3 \times \frac{\sqrt{2}}{2} \approx 3 \times 0.70710678118 = 2.121320 f(x3)=3cos(3π8)f(x_3) = 3 \cos\left(\frac{3\pi}{8}\right) f(x4)=3cos(π2)=3×0=0f(x_4) = 3 \cos\left(\frac{\pi}{2}\right) = 3 \times 0 = 0 f(x5)=3cos(5π8)f(x_5) = 3 \cos\left(\frac{5\pi}{8}\right)

Step 4: Set up the Riemann sum.

The Riemann sum is given by the formula:

S=i=0n1f(xi)ΔxS = \sum_{i=0}^{n-1} f(x_i) \Delta x

Substituting in the values:

S=Δx[f(x0)+f(x1)+f(x2)+f(x3)+f(x4)+f(x5)]S = \Delta x \left[ f(x_0) + f(x_1) + f(x_2) + f(x_3) + f(x_4) + f(x_5) \right]

S=π8[3+3cos(π8)+2.121320+3cos(3π8)+0+3cos(5π8)]S = \frac{\pi}{8} \left[ 3 + 3 \cos\left(\frac{\pi}{8}\right) + 2.121320 + 3 \cos\left(\frac{3\pi}{8}\right) + 0 + 3 \cos\left(\frac{5\pi}{8}\right) \right]

Now, we need to compute the cosine values for the remaining points. I will do that next.

Let me calculate that quickly. math

Given values

pi = math.pi

Function f(x) = 3 * cos(x)

def f(x): return 3 * math.cos(x)

Left endpoints for x

x_values = [0, pi/8, pi/4, 3pi/8, pi/2, 5pi/8]

Calculate f(x) at each left endpoint

f_values = [f(x) for x in x_values]

Riemann sum calculation

delta_x = pi / 8 riemann_sum = delta_x * sum(f_values) riemann_sumThe Riemann sum for the function f(x)=3cos(x)f(x) = 3 \cos(x) over the interval 0x3π40 \leq x \leq \frac{3\pi}{4}, using n=6n = 6 subintervals and left endpoints as sample points, is approximately:

S3.099558S \approx 3.099558

This value is rounded to six decimal places.

Would you like further explanation or have any other questions?

Here are some follow-up questions that could deepen the understanding:

  1. How would the Riemann sum change if we used right endpoints instead of left endpoints?
  2. What is the difference between a left Riemann sum and a right Riemann sum?
  3. How does the number of subintervals (nn) affect the accuracy of the Riemann sum?
  4. What is the exact value of the integral of f(x)=3cos(x)f(x) = 3 \cos(x) over this interval?
  5. How can we approximate the area under the curve using a larger value of nn?

Tip: Increasing the number of subintervals (nn) will generally improve the accuracy of the Riemann sum, bringing it closer to the exact integral value.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Riemann Sum
Trigonometric Functions
Cosine Function
Integral Approximation

Formulas

Riemann sum formula: S = Σ f(x_i) Δx
f(x) = 3 cos(x)

Theorems

Riemann Sum Theorem

Suitable Grade Level

Grades 11-12