Use a Maclaurin expansion to approximate $\int_{0}^{1/2}e^{-x^{2}}dx$ correct to within 0.001.

Solution Replace $x$ by $-x^{2}$ in the Maclaurin expansion for $e^{x}$ to obtain the Maclaurin expansion for $e^{-x^{2}}$. $$\begin{eqnarray*} e^{x} &=& 1+x+\dfrac{x^{2}}{2!}+\dfrac{x^{3}}{3!}+\,...\, \\[7pt] e^{-x^{2}} &=& 1-x^{2}+\frac{x^{4}}{2!}-\frac{x^{6}}{3!}+\frac{x^{8}}{4!}-\cdots \end{eqnarray*}$$

Now use the integration property of power series. $$\begin{eqnarray*} \int_{0}^{1/2}e^{-x^{2}}dx &=& \int_{0}^{1/2}\left( 1-x^{2}+\frac{x^{4}}{2!}-\frac{x^{6}}{3!}+\cdots \right)\,dx \\[9pt] &=& \left[ x-\frac{x^{3}}{3}+\frac{x^{5}}{2!5}-\frac{x^{7}}{3!7}+\cdots \right] _{0}^{1/2} \\[9pt] &=& \frac{1}{2}-\frac{1}{3(2)^{3}}+\frac{1}{2!5(2)^{5}}-\frac{1}{3!7(2)^{7}}+\cdots \\[9pt] &\approx & 0.5-0.041666+0.003125-0.000186+\cdots \end{eqnarray*}$$

Since this series satisfies the two conditions of the Alternating Series Test, the error due to using the first three terms as an approximation is less than the 4th term,

628

$\dfrac{1}{3!7(2)^{7}}= 1.\, 860119048\times 10^{-4}.$ So, $$\int_{0}^{1/2}e^{-x^{2}}dx\approx \frac{1}{2}-\frac{1}{3(2)^{3}}+\frac{1}{2!5(2)^{5}}\approx 0.461458$$

correct to within $0.001$.