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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 8, Problem 8.PE.101

Evaluate the integrals in Exercises 69–134. The integrals are listed in random order so you need to decide which integration technique to use.
∫ x dx / √(2 − x)

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1
Identify the integral to solve: \(\int \frac{x}{\sqrt{2 - x}} \, dx\).
Consider a substitution to simplify the integral. Let \(u = 2 - x\), which implies \(du = -dx\) and \(x = 2 - u\).
Rewrite the integral in terms of \(u\): replace \(x\) with \(2 - u\) and \(dx\) with \(-du\), so the integral becomes \(\int \frac{2 - u}{\sqrt{u}} (-du)\).
Distribute the negative sign and split the integral into two simpler integrals: \(-\int \frac{2}{\sqrt{u}} \, du + \int \frac{u}{\sqrt{u}} \, du\).
Simplify the integrands by expressing powers of \(u\) explicitly, then integrate each term using the power rule for integrals.

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Key Concepts

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

Integration by Substitution

Integration by substitution is a method used to simplify integrals by changing variables. It involves identifying a part of the integrand whose derivative also appears in the integral, allowing the integral to be rewritten in terms of a new variable. This technique is especially useful when dealing with composite functions like √(2 − x).
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Substitution With an Extra Variable

Algebraic Manipulation of the Integrand

Algebraic manipulation involves rewriting the integrand to a more convenient form before integrating. This can include factoring, expanding, or expressing parts of the integrand in terms of a substitution variable. Proper manipulation helps in recognizing patterns and applying the correct integration technique.
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Definite and Indefinite Integrals

Understanding the difference between definite and indefinite integrals is crucial. Indefinite integrals represent families of functions plus a constant of integration, while definite integrals compute the net area under a curve between two limits. This problem involves an indefinite integral, so the solution includes an arbitrary constant.
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