Q. 51 - Calculus

Question

Use Theorem 8.12 and the results from Exercises 41–50 to find series equal to the definite integrals in Exercises 51–60.

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

Step-by-Step Solution

Answer

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

Step 1. Given information is:

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

Step 2. Definite Integral

$$\begin{gathered} \mathrm{From} \mathrm{Q} 41. \\ \mathrm{Maclaurin} \mathrm{series} \mathrm{for} \mathrm{f}\left(\mathrm{x}\right)=\sin \left({\mathrm{x}}^3\right) \mathrm{is} \\ \sin \left({\mathrm{x}}^3\right)=\sum _{\mathrm{k}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{k}}}{\left(2\mathrm{k}+1\right)!}{\mathrm{x}}^{6\mathrm{k}+3} \\ \mathrm{Also}, \mathrm{F}=\int \mathrm{f} \\ \mathrm{F}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{k}}}{\left(2\mathrm{k}+1\right)!}\left(\frac{{\mathrm{x}}^{6\mathrm{k}+4}}{6\mathrm{k}+4}\right) \\ \mathrm{Adding} \mathrm{the} \mathrm{limits}, \\ \mathrm{F}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{k}}}{\left(2\mathrm{k}+1\right)!}{\left[\frac{{\mathrm{x}}^{6\mathrm{k}+4}}{6\mathrm{k}+4}\right]}_0^2 \\ =\sum _{\mathrm{k}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{k}}}{\left(2\mathrm{k}+1\right)!}\left[\frac{2^{6\mathrm{k}+4}-0}{6\mathrm{k}+4}\right] \\ =\sum _{\mathrm{k}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{k}}}{\left(2\mathrm{k}+1\right)!}\left[\frac{2^{6\mathrm{k}+4}}{6\mathrm{k}+4}\right] \\ =\sum _{\mathrm{k}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{k}}}{\left(2\mathrm{k}+1\right)!}\left[\frac{2^{6\mathrm{k}+3}}{3\mathrm{k}+2}\right] \end{gathered}$$