Evaluate the following integral. 4 √2 S 0 dx 16-x²

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
Question

Evaluate the Following Integral.

Evaluate the following integral.

\[
\int_{0}^{\frac{4}{\sqrt{2}}} \frac{dx}{\sqrt{16 - x^2}}
\]

---

Which substitution transforms the given integral into one that can be evaluated directly in terms of \(\theta\)?

- A. \(x = 4 \tan \theta\)
- B. \(x = 4 \sec \theta\)
- C. \(x = 4 \sin \theta\) (selected)

For this substitution, \(dx = (4 \cos (\theta)) \, d\theta\).

\[
\int_{0}^{\frac{4}{\sqrt{2}}} \frac{dx}{\sqrt{16 - x^2}} = \frac{4}{\sqrt{2}}
\]

(Type an exact answer, using \(\pi\) as needed.)
Transcribed Image Text:Evaluate the following integral. \[ \int_{0}^{\frac{4}{\sqrt{2}}} \frac{dx}{\sqrt{16 - x^2}} \] --- Which substitution transforms the given integral into one that can be evaluated directly in terms of \(\theta\)? - A. \(x = 4 \tan \theta\) - B. \(x = 4 \sec \theta\) - C. \(x = 4 \sin \theta\) (selected) For this substitution, \(dx = (4 \cos (\theta)) \, d\theta\). \[ \int_{0}^{\frac{4}{\sqrt{2}}} \frac{dx}{\sqrt{16 - x^2}} = \frac{4}{\sqrt{2}} \] (Type an exact answer, using \(\pi\) as needed.)
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Why do you add pi and subtract by (x) in step 2? Where does the pi come from? How does the integral go from 4/sqrt(2) to pi/4?

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