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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 8.1.65a

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
a. ∫(3/(x² + 4)) dx = ∫(3/x²) dx + ∫(3/4) dx.

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Recall the property of integrals that states \( \int (f(x) + g(x)) \, dx = \int f(x) \, dx + \int g(x) \, dx \). This means the integral of a sum is the sum of the integrals, but this property does not apply to quotients or products inside the integral unless they are separated as sums.
Look carefully at the given integral: \( \int \frac{3}{x^2 + 4} \, dx \). The denominator is \( x^2 + 4 \), which is a sum inside the denominator, not a sum of two separate fractions.
The statement claims that \( \int \frac{3}{x^2 + 4} \, dx = \int \frac{3}{x^2} \, dx + \int \frac{3}{4} \, dx \). This would mean splitting the fraction \( \frac{3}{x^2 + 4} \) into \( \frac{3}{x^2} + \frac{3}{4} \), which is not algebraically correct because \( \frac{1}{a + b} \neq \frac{1}{a} + \frac{1}{b} \).
To verify, consider a counterexample by choosing a specific value of \( x \), for example \( x = 1 \), and compare \( \frac{3}{1^2 + 4} = \frac{3}{5} \) with \( \frac{3}{1^2} + \frac{3}{4} = 3 + \frac{3}{4} = \frac{15}{4} \). Since these are not equal, the equality inside the integral does not hold.
Therefore, the original statement is false because the integral of a quotient with a sum in the denominator cannot be split into the sum of integrals of separate fractions as claimed.

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Key Concepts

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

Linearity of Integration

Integration is a linear operation, meaning the integral of a sum is the sum of the integrals. However, this property applies only when the integrand is expressed as a sum of separate terms, not when terms are combined inside a single denominator.
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Linearization

Properties of Rational Functions

A rational function is a ratio of polynomials. Splitting an integral like ∫(3/(x² + 4)) dx into ∫(3/x²) dx + ∫(3/4) dx is incorrect because (x² + 4) is a single denominator, not a sum of separate fractions.
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Properties of Functions

Counterexamples in Integration

To test if an integral equality holds, substituting specific values or comparing derivatives can provide counterexamples. Here, evaluating or differentiating the proposed split shows it does not equal the original integral, disproving the statement.
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Integration by Parts for Definite Integrals
Related Practice
Textbook Question

91. [Use of Tech] Regions bounded by exponentials Let a > 0 and let R be the region bounded by the graph of y = e^(-a·x) and the x-axis

on the interval [b, ∞).

a. Find A(a,b), the area of R as a function of a and b.

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Textbook Question

Arc length of a parabola Let L(c) be the length of the parabola f(x) = x² from x = 0 to x = c, where c ≥ 0 is a constant.

a. Find an expression for L.

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Textbook Question

60. Two Methods

a. Evaluate ∫(x · ln(x²)) dx using the substitution u = x² and evaluating ∫(ln(u)) du.

Textbook Question

57. Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

a. If x = 4 tanθ, then cscθ = 4/x.

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.

a. Use the reduction formula ∫ from 0 to ∞ of x^p e^(-x) dx = p ∫ from 0 to ∞ of x^(p-1) e^(-x) dx for p = 1, 2, 3, ...

to show that Γ(p + 1) = p! (p factorial).

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Textbook Question

41-44. {Use of Tech} Nonuniform grids

Use the indicated methods to solve the following problems with nonuniform grids.

41. A curling iron is plugged into an outlet at time t = 0. Its temperature T in degrees Fahrenheit, assumed to be a continuous function that is strictly increasing and concave down on 0 ≤ t ≤ 120, is given at various times (in seconds) in the table.

a. Approximate (1/120)∫(0 to 120)T(t)dt in three ways using a left Riemann sum, using a right Riemann sum and using the Trapezoid Rule

Interpret the value of (1/120)∫(0 to 120)T(t)dt in the context of this problem.