Chapter 8: Q. 61 (page 702)

Use the results from Exercises 51–60 and Theorem 7.38 to approximate the values of the definite integrals in Exercises 61–70 to within 0.001 of their values.
$\int_{0}^{2} \sin (x^{3}) d x$

Short Answer

Expert verified

The approximate value is $5 \cdot 2647$ .

Step by step solution

Step 1. Given information .

Consider the given definite integral $\int_{0}^{2} \sin (x^{3}) d x$ .

Step 2. Using the result of sin x from question 51 and theorem 7·38 .

The result of $\sin x = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{(2 k + 1)} x^{(2 k + 1)}$

Theorem $7 \cdot 38$ - Let L be the sum of an alternating series satisfying the

hypotheses of the alternating series test. For any term Sn in the sequence of partial sums, $|L - S_{n}|$ $<$ $a_{n + 1}$ . Furthermore, the sign of the difference L − Sn is the sign of the coefficient of the term $a_{n + 1}$ .

Step 3. Find the value .

$\int_{0}^{2} \sin (x^{3}) d x = \int_{0}^{2} \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{(2 k + 1)} {(x^{3})}^{(2 k + 1)} d x = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{(2 k + 1)} \int_{0}^{2} {(x^{3})}^{(2 k + 1)} d x = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{(2 k + 1)} \int_{0}^{2} x^{6 k + 1} d x = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{(2 k + 1)} {[\frac{x^{6 k + 3}}{6 k + 4}]}_{0}^{2} = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{(2 k + 1)} (\frac{2^{6 k + 1}}{3 k + 2})$

Substitute $k = 0, 1, 2, 3 . . . . . . . . \infty$

$\Rightarrow 1 - \frac{64}{15} + \frac{128}{15} - \frac{8}{3465} + . . . . . . . . . . . . . . . . . . \Rightarrow 5 \cdot 2647$

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Use the results from Exercises 51–60 and Theorem 7.38 to approximate the values of the definite integrals in Exercises 61–70 to within 0.001 of their values.
$\int_{0}^{2} \sin (x^{3}) d x$

Short Answer

Step by step solution

Step 1. Given information .

Step 2. Using the result of sin x from question 51 and theorem 7·38 .

Step 3. Find the value .

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Use the results from Exercises 51–60 and Theorem 7.38 to approximate the values of the definite integrals in Exercises 61–70 to within 0.001 of their values.∫02sinx3dx

Short Answer

Step by step solution

Step 1. Given information .

Step 2. Using the result of sin x from question 51 and theorem 7·38 .

Step 3. Find the value .

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

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Use the results from Exercises 51–60 and Theorem 7.38 to approximate the values of the definite integrals in Exercises 61–70 to within 0.001 of their values.
$\int_{0}^{2} \sin (x^{3}) d x$