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Chapter 5: Q. 21 (page 478)

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

Short Answer

Expert verified

The value of the integral is $3$ .

Step by step solution

01

Step 1. Given information.

The given improper integral is following,

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

02

Step 2. Value of integral.

Solving the integral,

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

03

Step 3. Value of integral.

If $B 鈫� 鈭�$ then $- \frac{3}{\sqrt[3]{B}} 鈫� 0$ ,

Therefore,

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

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Most popular questions from this chapter

Solve the integral: $鈭� \frac{3 x + 1}{s e c x} d x$

Suppose $u (x) = x^{2}$ . Calculate and compare the values of the following definite integrals:

role="math" localid="1648786835678" ${鈭�}_{- 1}^{5} u^{2} d u, {鈭�}_{x = - 1}^{x = 5} u^{2} d u and {鈭�}_{u (- 1)}^{u (5)} u^{2} d u$

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.

$x^{2} 鈭� 4 x 鈭� 8$

Solve given definite integral.

${鈭�}_{鈭� 1}^{1} 鈥� x^{3} \sqrt{9 x^{2} 鈭� 1} d x$

Solve the integral: $鈭� x \ln (x) d x$

See all solutions

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