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.

${\int }_0^2\sin \left(x^3\right)dx$

Short answer

The approximate value is$5·2647$.

Step-by-step solution

Step 1. Given information .

Consider the given definite integral${\int }_0^2\sin \left(x^3\right)dx$.

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

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

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,$\left|L-S_n\right|$$<$$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 .

$$\begin{gathered} {\int }_0^2\sin \left(x^3\right)dx={\int }_0^2\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k+1\right)}{\left(x^3\right)}^{\left(2k+1\right)}dx \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k+1\right)}{\int }_0^2{\left(x^3\right)}^{\left(2k+1\right)}dx \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k+1\right)}{\int }_0^2{x}^{6k+1}dx \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k+1\right)}{\left[\frac{x^{6k+3}}{6k+4}\right]}_0^2 \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k+1\right)}\left(\frac{2^{6k+1}}{3k+2}\right) \end{gathered}$$

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

$$\begin{gathered} \Rightarrow 1-\frac{64}{15}+\frac{128}{15}-\frac{8}{3465}+.................. \\ \Rightarrow 5·2647 \end{gathered}$$