Math Problem Statement

Estimate the integral \( \int_0^1 12 \cos^2(x) \ dx \) using the Midpoint Rule with \( n = 4 \) subintervals.

Solution

We are tasked with estimating the integral of 12cos⁡2(x)12 \cos^2(x) from 0 to 1 using the Midpoint Rule with n=4n = 4 subintervals.

Step-by-Step Solution:

1. Determine the interval width Δx\Delta x:

The interval is from a=0a = 0 to b=1b = 1, and the formula for the interval width is:

Δx=b−an=1−04=0.25\Delta x = \frac{b - a}{n} = \frac{1 - 0}{4} = 0.25

2. Find the midpoints xˉi\bar{x}_i:

For each subinterval, the midpoint is given by:

xˉi=xi−1+xi2\bar{x}_i = \frac{x_{i-1} + x_i}{2}

The endpoints of the subintervals are:

  • x0=0x_0 = 0
  • x1=0.25x_1 = 0.25
  • x2=0.5x_2 = 0.5
  • x3=0.75x_3 = 0.75
  • x4=1x_4 = 1

The midpoints will be:

\bar{x}_2 = \frac{0.25 + 0.5}{2} = 0.375, \quad \bar{x}_3 = \frac{0.5 + 0.75}{2} = 0.625, \quad \bar{x}_4 = \frac{0.75 + 1}{2} = 0.875$$ #### 3. **Evaluate the function at the midpoints:** We now evaluate the function $$f(x) = 12 \cos^2(x)$$ at the midpoints. - $$f(\bar{x}_1) = 12 \cos^2(0.125)$$ - $$f(\bar{x}_2) = 12 \cos^2(0.375)$$ - $$f(\bar{x}_3) = 12 \cos^2(0.625)$$ - $$f(\bar{x}_4) = 12 \cos^2(0.875)$$ #### 4. **Apply the Midpoint Rule:** The Midpoint Rule is given by: $$M_n = \Delta x \left[ f(\bar{x}_1) + f(\bar{x}_2) + f(\bar{x}_3) + f(\bar{x}_4) \right]$$ Substituting $$\Delta x = 0.25$$, we get: $$M_4 = 0.25 \left[ 12 \cos^2(0.125) + 12 \cos^2(0.375) + 12 \cos^2(0.625) + 12 \cos^2(0.875) \right]$$ I will now calculate the numerical value for each term and sum them. Let's compute that next.The Midpoint Rule approximation for the integral of $$12 \cos^2(x)$$ from 0 to 1 with $$n = 4$$ subintervals is approximately: $$M_4 \approx 8.7565$$ The function values at the midpoints were: - $$f(0.125) \approx 11.8135$$ - $$f(0.375) \approx 10.3901$$ - $$f(0.625) \approx 7.8919$$ - $$f(0.875) \approx 4.9305$$ If you would like further details or have any questions, feel free to ask! Here are 5 related questions you can explore: 1. How does the accuracy of the Midpoint Rule compare to other numerical methods like the Trapezoidal Rule? 2. What is the exact value of the integral, and how close is the Midpoint Rule approximation? 3. How would the result change if we increased the number of subintervals? 4. What happens to the Midpoint Rule's approximation when we apply it to different types of functions (e.g., sine functions)? 5. How does cosine's periodicity affect the accuracy of numerical integration over different intervals? **Tip:** The accuracy of the Midpoint Rule improves as you increase the number of subintervals $$n$$.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Numerical Integration
Midpoint Rule
Trigonometry

Formulas

Midpoint Rule: M_n = \Delta x [f(\bar{x}_1) + f(\bar{x}_2) + ... + f(\bar{x}_n)]
Function: f(x) = 12 \cos^2(x)
Midpoint formula: \bar{x}_i = \frac{1}{2}(x_{i-1} + x_i)

Theorems

Numerical approximation using the Midpoint Rule

Suitable Grade Level

Grades 11-12

Related Recommendation

Midpoint Rule Approximation of ∫[0 to π/2] 2 cos^4(x) dx with n = 4

Midpoint Rule Approximation for Integral of cos(x^2) from 0 to 1

Approximate Integral of (1 + cos(x)) / x using Simpson's Rule

Estimating ∫[0 to 1] 12 cos(x^2) dx Using the Trapezoidal and Midpoint Rule

Approximating Area Using Midpoint Rule for sin²(πx/4) and cos²(πx/4)