Math Problem Statement

For questions 3, 4, and 5, you will use b  =  2. Enter the value of M6, correct to 3 decimals, into the answer box below.

For questions 3, 4, and 5, you will use b = 2. Enter the value of M6, correct to 3 decimals, into the answer box below.

Solution

To modify the given code to approximate the integral 02ex3dx\int_0^2 e^{-x^3} \, dx using the midpoint Riemann sum M6M_6, we need to make a few adjustments in the code. Specifically, we will:

  1. Set b=2b = 2 (as specified).
  2. Update n=6n = 6 for M6M_6 (to divide the interval into 6 subintervals).
  3. Change the function definition to ex3e^{-x^3}.

Here’s the modified code with these updates:

python
import numpy as np # Define parameters a = 0 b = 2 n = 6 # Set to 6 for M6 delta_x = (b - a) / n # Compute the base length of each rectangle # Define independent variable x = np.linspace(a, b, n + 1) # x-values at the left and right endpoints of each subinterval x_mid = x[:-1] + delta_x / 2 # Midpoints of each interval # Define function for the integral y_mid = np.exp(-x_mid**3) # Compute the midpoint Riemann sum MS = sum(y_mid) * delta_x # Riemann sum approximation # Output statement print("The midpoint sum M6 is:", round(MS, 3))

Explanation of Changes:

  • Setting b=2b = 2: This limits the integration to the interval [0,2][0, 2].
  • Setting n=6n = 6: This gives us 6 subintervals, as required for M6M_6.
  • Defining f(x)=ex3f(x) = e^{-x^3}: The integrand function is updated to match the problem.

This code will calculate M6M_6, the midpoint Riemann sum approximation for the given integral, and print the result rounded to three decimal places.

Would you like further clarification on any part of this code? Here are some related questions to deepen your understanding:

  1. How does changing nn affect the accuracy of a Riemann sum approximation?
  2. What is the purpose of using midpoint values in this calculation?
  3. How does the Riemann sum method relate to the actual integral value?
  4. What would happen if we used a different number of intervals (e.g., M10M_{10})?
  5. Can you explain the choice of exp(x3)\exp(-x^3) as the function in this code?

Tip: When adjusting the number of subintervals in numerical methods like Riemann sums, smaller subintervals (higher nn) generally yield more accurate approximations.

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Math Problem Analysis

Mathematical Concepts

Numerical Integration
Riemann Sum
Midpoint Approximation
Exponentials

Formulas

Midpoint Riemann Sum M_n = sum(f(x_mid_i) * delta_x) for i = 0 to n - 1
delta_x = (b - a) / n
f(x) = e^{-x^3}

Theorems

Fundamental Theorem of Calculus (applied in approximating integrals)

Suitable Grade Level

College-level Calculus or Advanced High School Calculus