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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243당신이 사용하는 게 아니라요?교과서 변경
모든 교과서Briggs 3rd EditionCh. 8 - Integration Techniques문제 8.9.50
8장, 문제 8.9.50

7–58. Improper integrals Evaluate the following integrals or state that they diverge.
50. ∫ (from 0 to 9) 1/(x - 1)¹ᐟ³ dx

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Identify the type of integral and check for any points of discontinuity or singularities within the interval of integration. Here, the integrand is \(\frac{1}{(x - 1)^{1/3}}\), and the interval is from 0 to 9. Notice that the function is undefined at \(x = 1\) because the denominator becomes zero there.
Since \(x = 1\) lies within the interval of integration, split the integral at this point to handle the improper integral properly. Write the integral as the sum of two integrals: \(\int_0^1 \frac{1}{(x - 1)^{1/3}} \, dx + \int_1^9 \frac{1}{(x - 1)^{1/3}} \, dx\).
Rewrite each integral as a limit approaching the point of discontinuity. For the first integral, express it as \(\lim_{t \to 1^-} \int_0^t \frac{1}{(x - 1)^{1/3}} \, dx\). For the second integral, express it as \(\lim_{s \to 1^+} \int_s^9 \frac{1}{(x - 1)^{1/3}} \, dx\).
Find the antiderivative of the integrand \(\frac{1}{(x - 1)^{1/3}}\). Recall that \(\int (x - a)^n \, dx = \frac{(x - a)^{n+1}}{n+1} + C\) for \(n \neq -1\). Here, \(n = -\frac{1}{3}\), so the antiderivative is \(\frac{(x - 1)^{2/3}}{\frac{2}{3}} + C = \frac{3}{2} (x - 1)^{2/3} + C\).
Evaluate each limit by substituting the antiderivative back into the definite integrals and taking the limits as \(t \to 1^-\) and \(s \to 1^+\). Determine whether these limits converge to finite values or diverge to infinity to conclude if the original integral converges or diverges.

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주요 개념

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Improper Integrals

Improper integrals involve integrals with infinite limits or integrands with discontinuities within the interval. To evaluate them, one must consider limits approaching the problematic points to determine convergence or divergence.
추천 영상:
11:11
Improper Integrals: Infinite Intervals

Integrals with Discontinuous Integrands

When the integrand has a discontinuity inside the integration interval, the integral is split at the discontinuity point. Each part is evaluated as a limit approaching the discontinuity to check if the integral converges.
추천 영상:
05:22
Completing the Square to Rewrite the Integrand

Behavior of Functions Near Singularities

Understanding how functions behave near points where they become unbounded or undefined is crucial. For example, functions like 1/(x - a)^p converge near x = a if p < 1, and diverge if p ≥ 1, guiding the evaluation of improper integrals.
추천 영상:
5:46
Graphs of Exponential Functions
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교과서 질문

64. (Use of Tech) Normal distribution of movie lengths

A study revealed that the lengths of U.S. movies are normally distributed with a mean of 110 minutes and a standard deviation of 22 minutes. This means that the fraction of movies with lengths between a and b minutes (with a < b) is given by the integral:

(1/(22√(2π))) ∫[a to b] e^(-((x-110)/22)²/2) dx.

What percentage of U.S. movies are between 1 hr and 1.5 hr long (60-90 min)?

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7–64. Integration review Evaluate the following integrals.

32. ∫ from 0 to 2 of x / (x² + 4x + 8) dx

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교과서 질문

9–40. Integration by parts Evaluate the following integrals using integration by parts.

40. ∫ e^√x dx

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7–84. Evaluate the following integrals.

25. ∫ [1 / (x√(1 - x²))] dx

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74. A secant reduction formula

Prove that for positive integers n ≠ 1,

∫ secⁿ x dx = (secⁿ⁻² x tan x)/(n − 1) + (n − 2)/(n − 1) ∫ secⁿ⁻² x dx.

(Hint: Integrate by parts with u = secⁿ⁻² x and dv = sec² x dx.)

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교과서 질문

9–61. Trigonometric integrals Evaluate the following integrals.

59. ∫ from 0 to π/2 of √(1 - cos2x) dx

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