Q. 61 - Calculus

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.

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

Step-by-Step Solution

Answer

The approximate value is $5·2647$ .

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

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