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Chapter 5: Q. 24 (page 478)

Use limits of definite integrals to calculate each of the improper integrals in Exercises.
${鈭�}_{0}^{1} \frac{1}{\sqrt[4]{x^{3}}} d x$

Short Answer

Expert verified

The improper integral on $[0, 1]$ converges.

role="math" localid="1648836199546" ${鈭�}_{0}^{1} \frac{1}{\sqrt[4]{x^{3}}} d x = 4$

Step by step solution

01

Step 1. Given information.

The given improper integral is following.

${鈭�}_{0}^{1} \frac{1}{\sqrt[4]{x^{3}}} d x$

02

Step 2. Value of integral.

Write the integral in the proper form of ${鈭�}_{0}^{1} \frac{1}{x^{p}} d x .$

${鈭�}_{0}^{1} \frac{1}{x^{p}} d x = {鈭�}_{0}^{1} \frac{1}{\sqrt[4]{x^{3}}} d x {鈭�}_{0}^{1} \frac{1}{x^{p}} d x = {鈭�}_{0}^{1} \frac{1}{x^{\frac{3}{4}}} d x p = \frac{3}{4} & \frac{3}{4} < 1$

The Fundamental Theorem of Calculus tells us that the improper integral on $[0, 1]$ converges.

${鈭�}_{0}^{1} \frac{1}{\sqrt[4]{x^{3}}} d x = \lim_{a 鈫� 0} {鈭�}_{a}^{1} \frac{1}{\sqrt[4]{x^{3}}} d x = \lim_{a 鈫� 0^{+}} {[4 x^{\frac{1}{4}}]}_{a}^{1} = \lim_{a 鈫� 0^{+}} (4 - 4 a^{\frac{1}{4}}) = 4$

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Most popular questions from this chapter

Which of the integrals that follow would be good candidates for trigonometric substitution? If a trigonometric substitution is a good strategy, name the substitution. If another method is a better strategy, explain that method.

$(a) 鈭� \frac{4 + x^{2}}{x} d x$ $(b) 鈭� \frac{x}{4 + x^{2}} d x$

role="math" localid="1648759296940" $(c) 鈭� \frac{x^{2}}{4 + x^{2}} d x$ $(d) 鈭� \frac{16 鈭� x^{4}}{4 + x^{2}} d x$

Domains and ranges of inverse trigonometric functions: For each function that follows, (a) list the domain and range, (b) sketch a labeled graph, and (c) discuss the domains and ranges in the context of the unit circle.

$f (x) = \sec^{鈭� 1} 鈦� x$

Suppose you use polynomial long division to divide p(x) by q(x), and after doing your calculations you end up with the polynomial $x^{2} - x + 3$ as the quotient above the top line, and the polynomial 3x 鈭� 1 at the bottom as the remainder. Then $p (x) =___a n d \frac{p (x)}{q (x)} =____$

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

Solve the integral

$鈭� \sqrt{3 - x} \sqrt{x - 1} d x$

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