Question

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

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

Short 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-by-step solution

Step 1. Given information is: 

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

Step 2. Definite integral

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