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Chapter 8: Q. 52 (page 701)

Use Theorem 8.12 and the results from Exercises 41鈥�50 to find series equal to the definite integrals in Exercises 51鈥�60.
${鈭�}_{0}^{1} x \cos (x^{3}) dx$

Short Answer

Expert verified

${鈭�}_{0}^{1} x \cos (x^{3}) dx = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k)!} [\frac{1}{6 k + 2}]$

Step by step solution

01

Step 1. Given information is:

${鈭�}_{0}^{1} x \cos (x^{3}) d x$

02

Step 2. Definite integral

$From Q 42 . Maclaurin series for f (x) = x \cos (x^{3}) is x \cos {(x)}^{3} = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k)!} x^{6 k + 1} Also, F = 鈭� f F (x) = = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k)!} (\frac{x^{6 k + 2}}{6 k + 2}) Adding the limits, F (x) = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k)!} {[\frac{x^{6 k + 2}}{6 k + 2}]}_{0}^{1} = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k)!} [\frac{1^{6 k + 2} - 0}{6 k + 2}] = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k)!} [\frac{1}{6 k + 2}]$

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Most popular questions from this chapter

Let $f$ be a function with an nth-order derivative at a point $x_{o}$ and let $P_{n} (x) = {鈭�}_{k = 0}^{n} \frac{f^{(k)} (x_{0})}{k!} {(x - x_{0})}^{k}$ . Prove that $f^{(k)} (x_{0}) = P_{n}^{(k)} (x_{0})$ for every non-negative integer $k 鈮� n$ .

In Exercises 41鈥�48 find the fourth Taylor polynomial $P_{4} (x)$ for the specified function and the given value of $x_{0}$

$46 . \sqrt[3]{x}, 1$

If m is a positive integer, how can we find the Maclaurin series for the function $c x^{m} f (x)$ if we already know the Maclaurin series for the function f(x)? How do you find the interval of convergence for the new series?

Find the interval of convergence for power series: ${鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k + 1)!} {(3 x + 7)}^{2 k + 1}$

Find the interval of convergence for power series: ${鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k + 1)!} x^{2 k + 1}$

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