Q. 58 - 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. 

${\int }_{0.5}^2{\mathrm{x}}^3 \cos \left(\frac{\mathrm{x}}{2}\right) \mathrm{dx}$

Step-by-Step Solution

Answer

${\int }_{0.5}^2{\mathrm{x}}^3 \cos \left(\frac{\mathrm{x}}{2}\right) \mathrm{dx}=\sum _{\mathrm{k}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{k}}}{\left(2\mathrm{k}\right)!}{\left(\frac{1}{2}\right)}^{2\mathrm{k}+3}\left[\frac{2^{2\mathrm{k}+4}-{(0.5)}^{2\mathrm{k}+4}}{2\mathrm{k}+4}\right]$

Step 1. Given information is:

${\int }_{0.5}^2{\mathrm{x}}^3 \cos \left(\frac{\mathrm{x}}{2}\right) \mathrm{dx}$

Step 2. Definite integral

$$\begin{gathered} \mathrm{From} \mathrm{Q} 48. \\ \mathrm{Maclaurin} \mathrm{series} \mathrm{for} \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3 \cos \left(\frac{\mathrm{x}}{2}\right) \mathrm{is} \\ {\mathrm{x}}^3\cos \left(\frac{\mathrm{x}}{2}\right)=\sum _{\mathrm{k}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{k}}}{\left(2\mathrm{k}\right)!}{\left(\frac{\mathrm{x}}{2}\right)}^{2\mathrm{k}+3} \\ \mathrm{Also}, \mathrm{F}=\int \mathrm{f} \\ \mathrm{F}\left(\mathrm{x}\right) =\sum _{\mathrm{k}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{k}}}{\left(2\mathrm{k}\right)!}{\left(\frac{1}{2}\right)}^{2\mathrm{k}+3}\left(\frac{{\mathrm{x}}^{2\mathrm{k}+4}}{2\mathrm{k}+4}\right) \\ \mathrm{Adding} \mathrm{the} \mathrm{limits}, \\ \mathrm{F}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{k}}}{\left(2\mathrm{k}\right)!}{\left(\frac{1}{2}\right)}^{2\mathrm{k}+3}{\left[\frac{{\mathrm{x}}^{2\mathrm{k}+4}}{2\mathrm{k}+4}\right]}_{0.5}^2 \\ =\sum _{\mathrm{k}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{k}}}{\left(2\mathrm{k}\right)!}{\left(\frac{1}{2}\right)}^{2\mathrm{k}+3}\left[\frac{2^{2\mathrm{k}+4}-{(0.5)}^{2\mathrm{k}+4}}{2\mathrm{k}+4}\right] \end{gathered}$$