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Chapter 5: Q. 17 (page 464)

Why doesn’t the definite integral $\int_{2}^{3} \sqrt{1 - x^{2}} d x$ make sense? (Hint: Think about domains.)

Short Answer

Expert verified

Ans: $\sqrt{1 - x^{2}}$ is not defined in $[2, 3]$ .

Step by step solution

01

Step 1. Given information:

$\int_{2}^{3} \sqrt{1 - x^{2}} d x$

02

Step 2. Proving:

The definite integral $\int_{2}^{3} \sqrt{1 - x^{2}}$ does not make sense because the function has a domain in $[- 1, 1]$ . But the limit of the integral is $[2, 3]$ which does not come under the domain of the function. That's why $\sqrt{1 - x^{2}}$ is not defined in $[2, 3]$ .

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Most popular questions from this chapter

Solve the integral: $\int \ln (3 x) d x$

Solve the integral: $\int \frac{3 x + 1}{x} d x$

Problem Zero: Read the section and make your own summary of the material.

Explain why, if $x = a \tan u$ , then $\sqrt{x^{2} + a^{2}} = a s e c u$ . Your explanation should include a discussion of domains and absolute values.

True/False: Determine whether each of the statements that follow is true or false. If a statement is true, explain why. If a statement is false, provide a counterexample.

(a) True or False: The substitution x = 2 sec u is a suitable choice for solving $\int \frac{1}{x^{2} - 4} d x$ .

(b) True or False: The substitution x = 2 sec u is a suitable choice for solving $\int \frac{1}{\sqrt{x^{2} - 4}} d x$ .

(c) True or False: The substitution x = 2 tan u is a suitable choice for solving $\int \frac{1}{x^{2} + 4} d x .$

(d) True or False: The substitution x = 2 sin u is a suitable choice for solving $\int {(x^{2} + 4)}^{- 5 / 2} d x$

(e) True or False: Trigonometric substitution is a useful strategy for solving any integral that involves an expression of the form $\sqrt{x^{2} - a^{2}}$ .

(f) True or False: Trigonometric substitution doesn’t solve an integral; rather, it helps you rewrite integrals as ones that are easier to solve by other methods.

(g) True or False: When using trigonometric substitution with $x = a \sin u$ , we must consider the cases $x > a$ and $x < - a$ separately.

(h) True or False: When using trigonometric substitution with $x = a s e c u$ , we must consider the cases $x > a$ and $x < - a$ separately.

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