Q. 57 - 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. 

${\int }_1^2{\mathrm{x}}^2 \sin \left(5{\mathrm{x}}^2\right) \mathrm{dx}$

Step-by-Step Solution

Answer

${\int }_1^2{\mathrm{x}}^2 \sin \left(5{\mathrm{x}}^2\right) \mathrm{dx}==\sum _{\mathrm{k}=0}^{\infty } \frac{{\left(-1\right)}^{\mathrm{k}}}{\left(2\mathrm{k}+1\right)!}{5}^{2\mathrm{k}+1}\left[\frac{2^{4\mathrm{k}+5}-1}{4\mathrm{k}+5}\right]$

Step 1. Given information is:

${\int }_1^2{\mathrm{x}}^2 \sin \left(5{\mathrm{x}}^2\right) \mathrm{dx}$

Step 2. Definite integral

$$\begin{gathered} \mathrm{From} \mathrm{Q} 47. \\ \mathrm{Maclaurin} \mathrm{series} \mathrm{for} \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2 \sin \left(5{\mathrm{x}}^2\right) \mathrm{is} \\ {\mathrm{x}}^2\sin \left(5{\mathrm{x}}^2\right)=\sum _{\mathrm{k}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{k}}}{\left(2\mathrm{k}+1\right)!}{5}^{2\mathrm{k}+1} {\mathrm{x}}^{4\mathrm{k}+4} \\ \mathrm{Also}, \mathrm{F}=\int \mathrm{f} \\ \mathrm{F}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\infty } \frac{{\left(-1\right)}^{\mathrm{k}}}{\left(2\mathrm{k}+1\right)!}{5}^{2\mathrm{k}+1}\frac{{\mathrm{x}}^{4\mathrm{k}+5}}{4\mathrm{k}+5} \\ \mathrm{Adding} \mathrm{the} \mathrm{limits}, \\ \mathrm{F}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\infty } \frac{{\left(-1\right)}^{\mathrm{k}}}{\left(2\mathrm{k}+1\right)!}{5}^{2\mathrm{k}+1}{\left[\frac{{\mathrm{x}}^{4\mathrm{k}+5}}{4\mathrm{k}+5}\right]}_1^2 \\ =\sum _{\mathrm{k}=0}^{\infty } \frac{{\left(-1\right)}^{\mathrm{k}}}{\left(2\mathrm{k}+1\right)!}{5}^{2\mathrm{k}+1}\left[\frac{2^{4\mathrm{k}+5}-1^{4\mathrm{k}+5}}{4\mathrm{k}+5}\right] \\ =\sum _{\mathrm{k}=0}^{\infty } \frac{{\left(-1\right)}^{\mathrm{k}}}{\left(2\mathrm{k}+1\right)!}{5}^{2\mathrm{k}+1}\left[\frac{2^{4\mathrm{k}+5}-1}{4\mathrm{k}+5}\right] \end{gathered}$$