Question

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.

${\int }_0^{1}x\cos \left(x^3\right)dx$

Short answer

The approximate value is$0·06752$.

Step-by-step solution

Step 1. Given information .

Consider the given integral${\int }_{}^{1}x\cos \left(x^3\right)dx$.

Step 2. Using the result from Exercises 51–60 and Theorem 7.38 .

The result of $\cos x=\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k\right)}{x}^{2k}$

Theorem 7.38 - 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,. Furthermore, the sign of the difference L − Sn is the sign of the coefficient of the term .

Step 3. Find the value .

$$\begin{gathered} {\int }_0^{1}x\cos \left(x^3\right)dx={\int }_0^{1}x·\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k\right)}{\left(x^3\right)}^{2k}dx \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k\right)}{\int }_0^{1}x{x}^{6k}dx \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k\right)}{\int }_0^1{x}^{6k+1}dx \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k\right)}{\left[\frac{x^{6k+2}}{6k+2}\right]}_0^1 \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k\right)}\left[\frac{1}{6k+2}\right] \end{gathered}$$

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

$$\begin{gathered} \Rightarrow \frac{1}{8}-\frac{1}{16}+\frac{1}{192}-\frac{1}{5760}+............ \\ \Rightarrow 0·125-0·0625+0·00520-0·0001736 \\ \Rightarrow 0·06752 \end{gathered}$$