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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 8, Problem 8.8.66a

66–71. {Use of Tech} Estimating error Refer to Theorem 8.1 in the following exercises.
66. Let f(x) = cos(x²).
a. Find a Midpoint Rule approximation to ∫[-1 to 1] cos(x²) dx using n = 30 subintervals.

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1
Identify the interval of integration and the number of subintervals: here, the integral is from \(-1\) to \(1\), and \(n = 30\) subintervals.
Calculate the width of each subinterval using the formula \(\Delta x = \frac{b - a}{n}\), where \(a = -1\) and \(b = 1\).
Determine the midpoints of each subinterval. For the \(i^{th}\) subinterval, the midpoint \(x_i^*\) is given by \(x_i^* = a + \left(i - \frac{1}{2}\right) \Delta x\) for \(i = 1, 2, ..., n\).
Evaluate the function \(f(x) = \cos(x^2)\) at each midpoint \(x_i^*\) to get \(f(x_i^*)\).
Approximate the integral using the Midpoint Rule formula: \(M_n = \Delta x \sum_{i=1}^n f(x_i^*)\). This sum multiplied by the subinterval width gives the approximation.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Midpoint Rule for Numerical Integration

The Midpoint Rule approximates the definite integral of a function by dividing the interval into equal subintervals and using the function's value at each subinterval's midpoint to estimate the area. This method improves accuracy over using endpoints and is especially useful when the function is difficult to integrate analytically.
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Additional Rules for Indefinite Integrals

Partitioning the Interval into Subintervals

To apply numerical methods like the Midpoint Rule, the integration interval [a, b] is divided into n equal subintervals, each of width Δx = (b - a)/n. The midpoints of these subintervals are then used to evaluate the function, which forms the basis for the approximation.
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Error Estimation in Numerical Integration

Error estimation provides a bound on the difference between the exact integral and its numerical approximation. Theorem 8.1 likely gives an error bound for the Midpoint Rule, which depends on the second derivative of the function and the number of subintervals, helping assess the accuracy of the approximation.
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Determining Error and Relative Error