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Chapter 8: Q. 65 (page 702)

Use the results from Exercises 51鈥�60 and Theorem 7.38 to approximate the values of the definite integrals in Exercises 61鈥�70 to within 0.001 of their values.
${鈭�}_{0}^{0.3} \frac{dx}{1 + x^{3}}$

Short Answer

Expert verified

${鈭�}_{0}^{0.3} \frac{dx}{1 + x^{3}} 鈮� 0.297975$

Step by step solution

01

Step 1. Given information is:

${鈭�}_{0}^{0.3} \frac{dx}{1 + x^{3}}$

02

Step 2. Approximating the values

$From Q 55 . {鈭�}_{0}^{0.3} \frac{dx}{1 + x^{3}} = {鈭�}_{k = 0}^{鈭�} {(- 1)}^{k} [\frac{{(0.3)}^{3 k + 1}}{3 k + 1}] This implies that to approximate the values of integrals, put the value of k {鈭�}_{0}^{0.3} \frac{dx}{1 + x^{3}} = 0.3 - \frac{{(0.3)}^{4}}{4} + \frac{{(0.3)}^{7}}{7} - \frac{{(0.3)}^{10}}{10} + . . . = 0.3 - \frac{0.0081}{4} + \frac{0.002187}{7} 鈮� 0.297975$

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Most popular questions from this chapter

In Exercises 23鈥�32 we ask you to give Lagrange鈥檚 form for the corresponding remainder, $R_{4} (x)$

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${鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k} 蟺^{2 k + 1}}{(2 k + 1)!}$

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