Chapter 11 - Infinite Series - 11.7 Taylor Series - Exercises - Page 589: 51

$S = \mathop \smallint \limits_0^1 {{\rm{e}}^{ - {x^3}}}{\rm{d}}x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}}}{{\left( {3n + 1} \right)n!}}$ ${S_5} \approx 0.807446$

From Table 2: ${{\rm{e}}^x} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{x^n}}}{{n!}} = 1 + x + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $x$. Substituting $ - {x^3}$ for $x$ in the series above gives ${{\rm{e}}^{ - {x^3}}} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - {x^3}} \right)}^n}}}{{n!}} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{3n}}}}{{n!}}$, ${\ \ \ }$ converges for all $x$. Taking the definite integral on the series gives $\mathop \smallint \limits_0^1 {{\rm{e}}^{ - {x^3}}}{\rm{d}}x = \mathop \sum \limits_{n = 0}^\infty \mathop \smallint \limits_0^1 \dfrac{{{{\left( { - 1} \right)}^n}{x^{3n}}}}{{n!}}{\rm{d}}x = \mathop \sum \limits_{n = 0}^\infty \left[ {\dfrac{{{{\left( { - 1} \right)}^n}{x^{3n + 1}}}}{{\left( {3n + 1} \right)n!}}} \right]_0^1 = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}}}{{\left( {3n + 1} \right)n!}}$ Write $S = \mathop \smallint \limits_0^1 {{\rm{e}}^{ - {x^3}}}{\rm{d}}x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}}}{{\left( {3n + 1} \right)n!}}$. The alternating series above has positive terms: ${b_n} = \dfrac{1}{{\left( {3n + 1} \right)n!}}$. By Eq. (2) in Section 11.4: $\left| {S - {S_n}} \right| \lt {b_{n + 1}}$ We choose $n$ so that $\left| {S - {S_n}} \right| \lt {b_{n + 1}} \lt {10^{ - 4}}$ $\dfrac{1}{{\left( {3n + 4} \right)\left( {n + 1} \right)!}} \lt {10^{ - 4}}$ $\left( {3n + 4} \right)\left( {n + 1} \right)! \gt {10^4}$ We evaluate and list some values for $n$ in the following table: $\begin{array}{*{20}{c}} n&{}&{\left( {3n + 4} \right)\left( {n + 1} \right)!}\\ 3&{}&{312}\\ 4&{}&{1920}\\ 5&{}&{13680}\\ 6&{}&{110880} \end{array}$ From the results in the table above, we choose $N=5$ so that the series value is within an error of at most ${10^{ - 4}}$ to the definite integral. We compute the following: ${S_5} \approx 0.807446$, ${\ \ \ \ \ }$ $S \approx 0.807511$ It is verified that $\left| {S - {S_5}} \right| \approx 0.000065 \lt {10^{ - 4}}$.