Textbook Question
76. Different Substitutions
b. Show that ∫(1/√(x - x²)) dx = 2 sin⁻¹√x + C using substitution u = √x
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Ch. 8 - Integration Techniques
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 8, Problem 8.8.67a
66–71. {Use of Tech} Estimating error Refer to Theorem 8.1 in the following exercises.
67. Let f(x) = √(x³ + 1).
a. Find a Midpoint Rule approximation to ∫[1 to 6] √(x³ + 1) dx using n = 50 subintervals.
Verified step by step guidance1
Identify the integral to approximate: \(\int_{1}^{6} \sqrt{x^{3} + 1} \, dx\) and note that we will use the Midpoint Rule with \(n = 50\) subintervals.
Calculate the width of each subinterval using the formula \(\Delta x = \frac{b - a}{n}\), where \(a = 1\) and \(b = 6\). So, \(\Delta x = \frac{6 - 1}{50}\).
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 \(i = 1, 2, \ldots, 50\).
Evaluate the function \(f(x) = \sqrt{x^{3} + 1}\) at each midpoint \(x_i^*\) to get \(f(x_i^*)\).
Form the Midpoint Rule approximation by summing the function values at the midpoints multiplied by the subinterval width: \(M_n = \Delta x \sum_{i=1}^{50} f(x_i^*)\).
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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 complex or lacks an elementary antiderivative.
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Additional Rules for Indefinite Integrals
Partitioning the Interval into Subintervals
To apply the Midpoint Rule, the 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 helps in approximating the integral by summing the areas of rectangles centered at these midpoints.
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08:44Interval of Convergence
Error Estimation in Numerical Integration
Error estimation provides a bound on the difference between the exact integral and its numerical approximation. Theorem 8.1 typically gives an error bound for the Midpoint Rule involving the second derivative of the function, helping to assess the accuracy of the approximation and determine if the chosen number of subintervals is sufficient.
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04:57Determining Error and Relative Error
Related Practice
Textbook Question
63. Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
b. If m is a positive integer, then ∫[0 to π] sin^m(x) dx = 0.
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Textbook Question
85. Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
a. More than one integration method can be used to evaluate ∫ (1 / (1 - x²)) dx.
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Textbook Question
Computing areas On the interval [0,2], the graphs of f(x)=x²/3 and g(x)=x²(9−x²)^(-1/2) have similar shapes.
a. Find the area of the region bounded by the graph of f and the x-axis on the interval [0,2].
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Textbook Question
63. Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
a. If m is a positive integer, then ∫[0 to π] cos^(2m+1)(x) dx = 0.
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Textbook Question
Gamma function The gamma function is defined by Γ(p) = ∫ from 0 to ∞ of x^(p-1) e^(-x) dx, for p not equal to zero or a negative integer.
b. Use the substitution x = u² and the fact that ∫ from 0 to ∞ of e^(-u²) du = √(π/2) to show that Γ(1/2) = √π.
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