Textbook Question
5. e^(2t)-3e^t = 0
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 Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 7, Problem 7.6.84
Evaluate the integrals in Exercises 77–90.
84. ∫(from 2 to 4)2dx/(x²-6x+10)
Verified step by step guidance1
First, recognize that the integral is of the form \(\int_{2}^{4} \frac{2}{x^{2} - 6x + 10} \, dx\). The denominator is a quadratic expression, so start by completing the square for the quadratic \(x^{2} - 6x + 10\).
Rewrite the quadratic as \(x^{2} - 6x + 10 = (x^{2} - 6x + 9) + 1 = (x - 3)^{2} + 1\). This simplifies the integral to \(\int_{2}^{4} \frac{2}{(x - 3)^{2} + 1} \, dx\).
Make a substitution to simplify the integral: let \(u = x - 3\), so that \(du = dx\). Change the limits of integration accordingly: when \(x = 2\), \(u = 2 - 3 = -1\); when \(x = 4\), \(u = 4 - 3 = 1\).
Rewrite the integral in terms of \(u\): \(\int_{-1}^{1} \frac{2}{u^{2} + 1} \, du\). Recognize that the integral of \(\frac{1}{u^{2} + 1}\) is \(\arctan(u)\).
Use the integral formula \(\int \frac{1}{u^{2} + 1} \, du = \arctan(u) + C\) to write the integral as \(2 \int_{-1}^{1} \frac{1}{u^{2} + 1} \, du = 2 [\arctan(u)]_{-1}^{1}\). Evaluate this expression by substituting the limits.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Definite Integrals
A definite integral calculates the net area under a curve between two specific limits. It is represented as ∫ from a to b of f(x) dx, where a and b are the lower and upper 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
Integration of Rational Functions
Integrating rational functions often requires algebraic manipulation such as completing the square or partial fraction decomposition. Recognizing the form of the denominator helps in choosing the right technique to simplify the integral for easier evaluation.
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Completing the Square
Completing the square transforms a quadratic expression into a perfect square plus or minus a constant. This technique simplifies the denominator in integrals, making it easier to apply standard integral formulas involving inverse trigonometric functions or logarithms.
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Related Practice
Textbook Question
Find the limits in Exercises 13–20. (If in doubt, look at the function’s graph.)
17. lim(x→∞)arcsec(x)
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Textbook Question
Each of Exercises 1–4 gives a value of sinh x or cosh x. Use the definitions and the identity cosh²x - sinh²x = 1 to find the values of the remaining five hyperbolic functions.
4. cosh x = 13/5, x>0
Textbook Question
17. Show that √(10x+1) and √(x+1) grow at the same rate as x→∞ by showing that they both grow at the same rate as √x as x→∞.
Textbook Question
143.
b. Find the average value of ln(x) over [1, e].
Textbook Question
Evaluate the integrals in Exercises 53–76.
55. ∫dx/(17+x²)