Chapter 7: Problem 5
Approximate the integral using (a) the midpoint approximation \(M_{10},\) (b) the trapezoidal approximation \(T_{10}\), and (c) Simpson's rule approximation \(S_{20}\) using Formula (7). In each case, find the exact value of the integral and approximate the absolute error. Express your answers to at least four decimal places. $$ \int_{1}^{3} e^{-2 x} d x $$
Short Answer
Step by step solution
Define the integral and interval
Midpoint approximation \(M_{10}\)
Trapezoidal approximation \(T_{10}\)
Simpson's Rule approximation \(S_{20}\)
Exact value of the integral
Calculate the absolute errors
Summary of results
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Midpoint Approximation
In this exercise, we approximate the integral \( \int_{1}^{3} e^{-2x} \, dx \) using the midpoint approximation with 10 subintervals. Here's the breakdown:
- Calculate the width of each subinterval: \( \Delta x = \frac{3-1}{10} = 0.2 \).
- Determine the midpoint for each subinterval, given by \( x_i = 1 + 0.1 + i \cdot 0.2 \) for \( i = 0, 1, \ldots, 9 \).
- Evaluate the function \( e^{-2x} \) at each midpoint and sum up the results: \( M_{10} = 0.2 \sum_{i=0}^{9} e^{-2(1.2 + 0.2i)} \).
Trapezoidal Rule
For the integral \( \int_{1}^{3} e^{-2x} \, dx \) using the trapezoidal rule:
- The interval is divided into 10 subintervals with \( \Delta x = 0.2 \).
- The formula is \( T_{10} = \frac{\Delta x}{2} \left[ f(x_0) + 2\sum_{i=1}^{9} f(x_i) + f(x_{10}) \right] \).
- Plug in the endpoint values of each interval to compute: \( T_{10} \approx 0.0528 \).
Simpson's Rule
For our integral \( \int_{1}^{3} e^{-2x} \, dx \), Simpson’s Rule gives astonishing precision:
- Divide the interval into 20 subintervals, yielding \( \Delta x = 0.1 \).
- Utilize the formula: \( S_{20} = \frac{\Delta x}{3} \left[ f(x_0) + 4 \sum_{i=1,3,\ldots,19} f(x_i) + 2\sum_{i=2,4,\ldots,18} f(x_i) + f(x_{20}) \right] \).
- Calculate the sum using values at odd and even index points to achieve \( S_{20} \approx 0.0527 \).