Question

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

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

Short 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-by-step solution

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

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