Answer
$S = \mathop \smallint \limits_0^1 \cos \left( {{x^2}} \right){\rm{d}}x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}}}{{\left( {4n + 1} \right)\left( {2n} \right)!}}$
${S_3} \approx 0.904523$
Work Step by Step
From Table 2:
$\cos x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n}}}}{{\left( {2n} \right)!}} = 1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - \dfrac{{{x^6}}}{{6!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $x$.
Substituting ${x^2}$ for $x$ in the series above gives
$\cos \left( {{x^2}} \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{4n}}}}{{\left( {2n} \right)!}} = 1 - \dfrac{{{x^4}}}{{2!}} + \dfrac{{{x^8}}}{{4!}} - \dfrac{{{x^{12}}}}{{6!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $x$.
Taking the definite integral on the series gives
$\mathop \smallint \limits_0^1 \cos \left( {{x^2}} \right){\rm{d}}x = \mathop \sum \limits_{n = 0}^\infty \mathop \smallint \limits_0^1 \dfrac{{{{\left( { - 1} \right)}^n}{x^{4n}}}}{{\left( {2n} \right)!}}{\rm{d}}x$
$ = \mathop \sum \limits_{n = 0}^\infty \left[ {\dfrac{{{{\left( { - 1} \right)}^n}{x^{4n + 1}}}}{{\left( {4n + 1} \right)\left( {2n} \right)!}}} \right]_0^1 = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}}}{{\left( {4n + 1} \right)\left( {2n} \right)!}}$
Write $S = \mathop \smallint \limits_0^1 \cos \left( {{x^2}} \right){\rm{d}}x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}}}{{\left( {4n + 1} \right)\left( {2n} \right)!}}$.
The alternating series above has positive terms: ${b_n} = \dfrac{1}{{\left( {4n + 1} \right)\left( {2n} \right)!}}$.
By Eq. (2) in Section 11.4:
$\left| {S - {S_n}} \right| \lt {b_{n + 1}}$
We choose $n$ so that
$\left| {S - {S_n}} \right| \lt {b_{n + 1}} \lt {10^{ - 4}}$
$\dfrac{1}{{\left( {4n + 5} \right)\left( {2n + 2} \right)!}} \lt {10^{ - 4}}$
$\left( {4n + 5} \right)\left( {2n + 2} \right)! \gt {10^4}$
We evaluate and list some values for $n$ in the following table:
$\begin{array}{*{20}{c}}
n&{}&{\left( {4n + 5} \right)\left( {2n + 2} \right)!}\\
1&{}&{216}\\
2&{}&{9360}\\
3&{}&{685440}\\
4&{}&{76204800}
\end{array}$
From the results in the table above, we choose $N=3$ so that the series value is within an error of at most ${10^{ - 4}}$ to the definite integral.
We compute the following:
${S_3} \approx 0.904523$, ${\ \ \ \ \ }$ $S \approx 0.904524$
It is verified that $\left| {S - {S_3}} \right| \approx 0.00000145 \lt {10^{ - 4}}$.
