Textbook Question
Indeterminate Powers and Products
Find the limits in Exercises 53–68.
53. lim (x → 1⁺) x^(1/(1 - x))
Ch. 7 - Transcendental Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 7, Problem 7.3.89
Evaluate the integrals in Exercises 87–96.
89. ∫₀¹ 2^(−θ) dθ
Verified step by step guidance1
Identify the integral to be evaluated: \(\int_0^1 2^{-\theta} \, d\theta\).
Rewrite the integrand using the exponential function with base \(e\): \(2^{-\theta} = e^{-\theta \ln(2)}\).
Set up the integral with the exponential form: \(\int_0^1 e^{-\theta \ln(2)} \, d\theta\).
Integrate using the formula for the integral of an exponential function: \(\int e^{ax} \, dx = \frac{1}{a} e^{ax} + C\). Here, \(a = -\ln(2)\).
Evaluate the definite integral by substituting the limits \(\theta = 0\) and \(\theta = 1\) into the antiderivative and subtracting: \(\left[ \frac{e^{-\theta \ln(2)}}{-\ln(2)} \right]_0^1\).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Definite Integral
A definite integral calculates the net area under a curve between two limits. It is represented as ∫_a^b f(x) dx, where a and b are the bounds. Evaluating a definite integral involves finding the antiderivative and then applying the Fundamental Theorem of Calculus.
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Definition of the Definite Integral
Exponential Functions with Variable Exponents
Exponential functions like 2^(-θ) have a constant base and a variable exponent. Understanding how to integrate such functions requires rewriting them using natural logarithms or recognizing their derivative patterns, as the integral of a^x involves dividing by ln(a).
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Fundamental Theorem of Calculus
This theorem links differentiation and integration, stating that if F is an antiderivative of f, then ∫_a^b f(x) dx = F(b) - F(a). It allows evaluation of definite integrals by computing the difference of antiderivative values at the bounds.
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Related Practice
Textbook Question
In Exercises 7–38, find the derivative of y with respect to x, t, or θ, as appropriate.
10. y = ln(t^(3/2))
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Textbook Question
Evaluate the integrals in Exercises 77–90.
79. ∫(from -1 to 0)6dt/√(3-2t-t²)
Textbook Question
Solve the initial value problems in Exercises 115–120.
117. dy/dx = 1/(x√(x² - 1)), x > 1; y(2) = π
Textbook Question
In Exercises 7–38, find the derivative of y with respect to x, t, or θ, as appropriate.
23. y = ln(x)/(1+ln(x))
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Textbook Question
Use the results of Exercise 55 to show that the functions in Exercises 56–60 have inverses over their domains. Find a formula for df⁻¹/dx using Theorem 1.
f(x) = 27x³