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.

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

Short answer

${\int }_{0.5}^2{\mathrm{x}}^3\cos \left(\frac{\mathrm{x}}{2}\right)\mathrm{dx}\approx -0.342$

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. Approximating the values 

$$\begin{gathered} \mathrm{From}\mathrm{Q}58. \\ {\int }_{0.5}^2{\mathrm{x}}^3\cos \left(\frac{\mathrm{x}}{2}\right)\mathrm{dx}=\sum _{\mathrm{k}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{k}}}{\left(2\mathrm{k}\right)!}{\left(\frac{1}{2}\right)}^{2\mathrm{k}+3}\left[\frac{2^{2\mathrm{k}+4}-{(0.5)}^{2\mathrm{k}+4}}{2\mathrm{k}+4}\right] \\ \mathrm{This}\mathrm{implies}\mathrm{that}\mathrm{to}\mathrm{approximate}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{integrals},\mathrm{put}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{k} \\ {\int }_{0.5}^2{\mathrm{x}}^3\cos \left(\frac{\mathrm{x}}{2}\right)\mathrm{dx}=\frac{1}{2^3·4}\left(2^4-0.5^4\right)-\frac{1}{2!·2^5·6}\left(2^6-0.5^6\right)+\frac{1}{4!·2^7·8}\left(2^8-0.5^8\right) \\ =0.4980-0.1666+0.0104-0.000277+... \\ \approx -0.342 \end{gathered}$$