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Chapter 5: Q. 23 (page 478)

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

Short Answer

Expert verified

the improper integral on $[0, 1]$ diverges.

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

Step by step solution

01

Step 1. Given information.

The given improper integral is following.

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

02

Step 2. Value of integral.

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

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

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

localid="1648836050413" ${鈭�}_{0}^{1} \frac{1}{\sqrt[3]{x^{4}}} d x = \lim_{a 鈫� 0} {鈭�}_{a}^{1} \frac{1}{\sqrt[3]{x^{4}}} d x = \lim_{a 鈫� 0^{+}} {[- 3 x^{\frac{- 1}{3}}]}_{a}^{1} = \lim_{a 鈫� 0^{+}} (- 3 + 3 a^{\frac{- 1}{3}}) = - 鈭�$

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Most popular questions from this chapter

Why is it okay to use a triangle without thinking about the unit circle when simplifying expressions that result from a trigonometric substitution with $x = a \sin u$ or $x = a \tan u$ ? Why do we need to think about the unit circle after trigonometric substitution with $x = a s e c u$ ?

Complete the square for each quadratic in Exercises 28鈥�33. Then describe the trigonometric substitution that would be appropriate if you were solving an integral that involved that quadratic.

$2 {(x + 2)}^{2}$

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" $(c) 鈭� \frac{x^{2}}{4 + x^{2}} d x$ $(d) 鈭� \frac{16 鈭� x^{4}}{4 + x^{2}} d x$

Solve the integral

$鈭� \frac{1}{x^{2} \sqrt{x^{2} - 9}} d x$

Solve each of the integrals in Exercises 39鈥�74. Some integrals require trigonometric substitution, and some do not. Write your answers as algebraic functions whenever possible.

$鈭� \frac{\sqrt{4 - x^{2}}}{x^{2}} d x$

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