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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
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Tutti i libri di testoBriggs 3rd EditionCh. 8 - Integration TechniquesProblema 8.5.3c
Capitolo 8, Problema 8.5.3c

3. What term(s) should appear in the partial fraction decomposition of a proper rational function with each of the following?
c. A factor of (x² + 2x + 6) in the denominator

Guida verificata passo dopo passo
1
Identify that the factor given, \(x^2 + 2x + 6\), is an irreducible quadratic factor because its discriminant \(\Delta = b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot 6 = 4 - 24 = -20 < 0\), which means it cannot be factored further over the real numbers.
Recall that for an irreducible quadratic factor \(ax^2 + bx + c\) in the denominator, the corresponding term in the partial fraction decomposition has the form \(\frac{Ax + B}{ax^2 + bx + c}\), where \(A\) and \(B\) are constants to be determined.
If the factor \(x^2 + 2x + 6\) appears with multiplicity 1 (i.e., it is not repeated), then the partial fraction term is simply \(\frac{Ax + B}{x^2 + 2x + 6}\).
If the factor appears with multiplicity greater than 1, say \(n\), then the decomposition includes terms for each power from 1 up to \(n\): \(\frac{A_1x + B_1}{x^2 + 2x + 6} + \frac{A_2x + B_2}{(x^2 + 2x + 6)^2} + \cdots + \frac{A_nx + B_n}{(x^2 + 2x + 6)^n}\).
Summarize that the key point is that each irreducible quadratic factor in the denominator corresponds to a numerator that is a linear polynomial \(Ax + B\) over that quadratic factor in the partial fraction decomposition.

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Partial Fraction Decomposition

Partial fraction decomposition is a method used to express a proper rational function as a sum of simpler fractions whose denominators are factors of the original denominator. This technique simplifies integration and other operations by breaking complex fractions into manageable parts.
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Percorso guidato
10:07
Partial Fraction Decomposition: Distinct Linear Factors

Irreducible Quadratic Factors

An irreducible quadratic factor is a quadratic polynomial that cannot be factored further over the real numbers. In partial fraction decomposition, such factors in the denominator correspond to terms with linear numerators, for example, (Ax + B)/(quadratic factor).
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Percorso guidato
13:42
Partial Fraction Decomposition: Irreducible Quadratic Factors

Proper Rational Function

A proper rational function is a fraction where the degree of the numerator is less than the degree of the denominator. This condition ensures that partial fraction decomposition can be applied directly without polynomial division.
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Intro to Rational Functions
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45–48. {Use of Tech} Trapezoid Rule and Simpson’s Rule Consider the following integrals and the given values of n.

46. ∫(0 to 2) x⁴ dx; n = 30

c. Compute the absolute errors in the Trapezoid Rule and Simpson’s Rule with 2n subintervals.

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82. A family of exponentials The curves y = x * e^(-a * x) are shown in the figure for a = 1, 2, and 3.

c. Find the area of the region bounded by y = x * e^(-a * x) and the x-axis on the interval [0, b]. Because this area depends on a and b, we call it A(a, b).

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Gaussians An important function in statistics is the Gaussian (or normal distribution, or bell-shaped curve), f(x) = e^(-ax²).

c. Complete the square to evaluate ∫ from -∞ to ∞ of e^(-(ax² + bx + c)) dx, where a > 0, b, and c are real numbers.

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75. Exploring powers of sine and cosine

c. Prove that ∫₀ᵖⁱ sin²(nx) dx has the same value for all positive integers n.

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Prove the following orthogonality relations (which are used to generate Fourier series). Assume m and n are integers with m ≠ n.

c.

π

∫ sin(mx) cos(nx) dx = 0, when |m + n| is even

0

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94. [Use of Tech] Skydiving A skydiver has a downward velocity given by v(t) = V_T [(1 - e^(-2gt/V_T))/(1 + e^(-2gt/V_T))],

where t = 0 is the instant the skydiver starts falling, g = 9.8 m/s² is the acceleration due to gravity, and V_T is the terminal velocity of the skydiver.

c. Verify by integration that the position function is given by

s(t) = V_T t + (V_T²/g) ln[(1 + e^(-2gt/V_T))/2],

where s'(t) = v(t) and s(0) = 0.

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