1. The Maclaurin series for [latex]{e}^{\text{-}{x}^{2}}[/latex] is given by
   
   [latex]\begin{array}{cc}\hfill {e}^{\text{-}{x}^{2}}& ={\displaystyle\sum _{n=0}^{\infty}}\frac{{\left(\text{-}{x}^{2}\right)}^{n}}{n\text{!}}\hfill \\ & =1-{x}^{2}+\frac{{x}^{4}}{2\text{!}}-\frac{{x}^{6}}{3\text{!}}+\cdots +{\left(-1\right)}^{n}\frac{{x}^{2n}}{n\text{!}}+\cdots \hfill \\ & ={\displaystyle\sum _{n=0}^{\infty}}{\left(-1\right)}^{n}\frac{{x}^{2n}}{n\text{!}}.\hfill \end{array}[/latex]

   
   
   Therefore,
   

   [latex]\begin{array}{cc}\hfill {\displaystyle\int {e}^{\text{-}{x}^{2}}dx}& ={\displaystyle\int \left(1-{x}^{2}+\frac{{x}^{4}}{2\text{!}}-\frac{{x}^{6}}{3\text{!}}+\cdots +{\left(-1\right)}^{n}\frac{{x}^{2n}}{n\text{!}}+\cdots \right)dx}\hfill \\ & =C+x-\frac{{x}^{3}}{3}+\frac{{x}^{5}}{5.2\text{!}}-\frac{{x}^{7}}{7.3\text{!}}+\cdots +{\left(-1\right)}^{n}\frac{{x}^{2n+1}}{\left(2n+1\right)n\text{!}}+\cdots .\hfill \end{array}[/latex]

    

2. Using the result from part a. we have
   
   [latex]{\displaystyle\int }_{0}^{1}{e}^{\text{-}{x}^{2}}dx=1-\frac{1}{3}+\frac{1}{10}-\frac{1}{42}+\frac{1}{216}-\cdots [/latex].

   
   
   The sum of the first four terms is approximately [latex]0.74[/latex]. By the alternating series test, this estimate is accurate to within an error of less than [latex]\frac{1}{216}\approx 0.0046296<0.01[/latex].

Since [latex]\mu =100,\sigma =50[/latex], and we are trying to determine the area under the curve from [latex]a=100[/latex] to [latex]b=200[/latex], the integral becomes

The Maclaurin series for [latex]{e}^{\text{-}\frac{{x}^{2}}{2}}[/latex] is given by

Therefore,

Using the first five terms, we estimate that the probability is approximately [latex]0.4922[/latex]. By the alternating series test, we see that this estimate is accurate to within

We use the binomial series, replacing [latex]x[/latex] with [latex]\text{-}{k}^{2}{\sin}^{2}\theta [/latex]. Then we can write the period as

1. Using just the first term in the integrand, the first-order estimate is
   
   [latex]T\approx 4\sqrt{\frac{L}{g}}{\displaystyle\int }_{0}^{\frac{\pi}{2}}d\theta =2\pi \sqrt{\frac{L}{g}}[/latex].

   
   
   If [latex]{\theta }_{\text{max}}[/latex] is small, then [latex]k=\sin\left(\frac{{\theta }_{\text{max}}}{2}\right)[/latex] is small. We claim that when [latex]k[/latex] is small, this is a good estimate. To justify this claim, consider
   

   [latex]{\displaystyle\int }_{0}^{\frac{\pi}{2}}\left(1+\frac{1}{2}{k}^{2}{\sin}^{2}\theta +\frac{1\cdot 3}{2\text{!}{2}^{2}}{k}^{4}{\sin}^{4}\theta +\cdots \right)d\theta [/latex].

   
   
   Since [latex]|\sin{x}|\le 1[/latex], this integral is bounded by
   

   [latex]{\displaystyle\int }_{0}^{\frac{\pi}{2}}\left(\frac{1}{2}{k}^{2}+\frac{1.3}{2\text{!}{2}^{2}}{k}^{4}+\cdots \right)d\theta <\frac{\pi }{2}\left(\frac{1}{2}{k}^{2}+\frac{1\cdot 3}{2\text{!}{2}^{2}}{k}^{4}+\cdots \right)[/latex].

   
   
   Furthermore, it can be shown that each coefficient on the right-hand side is less than [latex]1[/latex] and, therefore, that this expression is bounded by
   

   [latex]\frac{\pi {k}^{2}}{2}\left(1+{k}^{2}+{k}^{4}+\cdots \right)=\frac{\pi {k}^{2}}{2}\cdot \frac{1}{1-{k}^{2}}[/latex],

   
   
   which is small for [latex]k[/latex] small.

2. For larger values of [latex]{\theta }_{\text{max}}[/latex], we can approximate [latex]T[/latex] by using more terms in the integrand. By using the first two terms in the integral, we arrive at the estimate
   
   [latex]\begin{array}{cc}\hfill T& \approx 4\sqrt{\frac{L}{g}}{\displaystyle\int }_{0}^{\frac{\pi}{2}}\left(1+\frac{1}{2}{k}^{2}{\sin}^{2}\theta \right)d\theta \hfill \\ & =2\pi \sqrt{\frac{L}{g}}\left(1+\frac{{k}^{2}}{4}\right).\hfill \end{array}[/latex]