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Chapter 8: Q. 70 (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.
${鈭�}_{0}^{0.3} x^{4} \tan^{- 1} (3 x^{3}) dx$

Short Answer

Expert verified

${鈭�}_{0}^{0.3} x^{4} \tan^{- 1} (3 x^{3}) dx 鈮� 0$

Step by step solution

01

Step 1. Given information is:

${鈭�}_{0}^{0.3} x^{4} \tan^{- 1} (3 x^{3}) dx$

02

Step 2. Approximating the values

$From Q 60 . {鈭�}_{0}^{0.3} x^{4} \tan^{- 1} (3 x^{3}) dx = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} {(3)}^{2 k + 1} [\frac{{(0.3)}^{6 k + 8}}{6 k + 8}] {鈭�}_{0}^{0.3} x^{4} \tan^{- 1} (3 x^{3}) dx = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} [\frac{{(3)}^{8 k + 9}}{10^{6 k + 8}}] This implies that to approximate the values of integrals, put the value of k {鈭�}_{0}^{0.3} x^{4} \tan^{- 1} (3 x^{3}) dx = \frac{3^{9}}{8 {(10)}^{8}} - \frac{3^{17}}{3 路 14 {(10)}^{14}} + . . . = 0.0000246 - . . . 鈮� 0$

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