Chapter 8.4, Problem 56E

a.

To determine

To graph: The function $f$ and find the intervals on which $f$ is increasing, the intervals on which $f$ is decreasing, and any local extreme values and where they occur if $f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{\frac{-{\left(x-\mu \right)}^2}{2{\sigma }^2}}\text{ let }\mu =0\text{ and }\sigma =1$ .

Explanation of Solution

Given information:

It is given that $f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{\frac{-{\left(x-\mu \right)}^2}{2{\sigma }^2}}$ and the given value are $\mu =0\text{ and }\sigma =1$ .

Graph:

Consider the given function:

  $f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{\frac{-{\left(x-\mu \right)}^2}{2{\sigma }^2}}$

Substitute the value $\mu =0\text{ and }\sigma =1$ ,

  $\begin{array}{l}f\left(x\right)=\frac{1}{\left(1\right)\sqrt{2\pi }}{e}^{\frac{-{\left(x-0\right)}^2}{2{\left(1\right)}^2}}\\ f\left(x\right)=\frac{1}{\sqrt{2\pi }}{e}^{\frac{-x^2}{2}}\end{array}$

Now use the graphing calculator to draw the graph of $f\left(x\right)=\frac{1}{\sqrt{2\pi }}{e}^{\frac{-x^2}{2}}$:

  

Interpretation: From the graph it can be seen that $f$ is increasing on $\left(-\infty ,0\right]$ , and $f$ is decreasing on $\left(0,\infty \right]$ . It has a local maxima at $\left(0,f\left(0\right)\right)=\left(0,\frac{1}{\sqrt{2\pi }}\right)$ .

b.

To determine

To state: The value of ${\displaystyle {\int }_{-n}^{n}f\left(x\right)dx}\text{ for }n=1,2,3$ .

Answer to Problem 56E

The resultant answer is $0.683,0.954,\text{ and }0.997\text{ for }n=\text{1,2,3}$ respectively.

Explanation of Solution

Given information:

From part (a) $f\left(x\right)=\frac{1}{\sqrt{2\pi }}{e}^{\frac{-x^2}{2}}$ .

Consider the integral ${\displaystyle {\int }_{-n}^n\frac{1}{\sqrt{2\pi }}{e}^{\frac{-x^2}{2}}}dx$ ,

Now evaluate it for the values of $n=1,2,3$ , then

  $\begin{array}{l}\text{NINT}\left(\frac{1}{\sqrt{2\pi }}{e}^{\frac{-x^2}{2}},x,-1,1\right)=0.683\\ \text{NINT}\left(\frac{1}{\sqrt{2\pi }}{e}^{\frac{-x^2}{2}},x,-2,2\right)=0.954\\ \text{NINT}\left(\frac{1}{\sqrt{2\pi }}{e}^{\frac{-x^2}{2}},x,-3,3\right)=0.997\end{array}$

Therefore, the values are $0.683,0.954,\text{ and }0.997\text{ for }n=\text{1,2,3}$ .

c.

To determine

To state: A convincing argument that ${\displaystyle {\int }_{-\infty }^{\infty }f\left(x\right)}dx=1$ .

Answer to Problem 56E

As $b\to \infty ,2e^{\frac{-b}{2}}\to 0$ , so for large enough b , ${\displaystyle {\int }_b^{\infty }f\left(x\right)dx}$ is as small as someone want. Likewise, for large enough b , ${\displaystyle {\int }_{-b}^{\infty }f\left(x\right)dx}$ is as small as a person wants.

Explanation of Solution

Given information:

The given statement says to give a convincing argument that ${\displaystyle {\int }_{-\infty }^{\infty }f\left(x\right)}dx=1$ .

Part (b) suggests that b increases the integral approaches 1. Then ${\displaystyle {\int }_{-b}^{b}f\left(x\right)dx}$ can be made as close to 1 as it is chosen that said that as $b>1$ large enough.

Also it can be made ${\displaystyle {\int }_{-\infty }^{b}f\left(x\right)dx}$ and ${\displaystyle {\int }_{-b}^{\infty }f\left(x\right)dx}$ as small as it is wanted by choosing b large enough. This is because $0<f\left(x\right)<e^{\frac{-x}{2}}\text{ for }x>1$ .

Thus for $b>1$ ,

  $\begin{array}{c}{\displaystyle {\int }_b^{\infty }f\left(x\right)dx}<{\displaystyle {\int }_b^{\infty }{e}^{\frac{-x}{2}}dx}\\ {\displaystyle {\int }_b^{\infty }{e}^{\frac{-x}{2}}dx}={}_c{\underrightarrow{\lim }}_{\infty }{\displaystyle {\int }_b^c{e}^{\frac{-x}{2}}dx}\\ ={}_c{\underrightarrow{\lim }}_{\infty }{\left[-2e^{\frac{-x}{2}}\right]}_b^c\\ ={}_c{\underrightarrow{\lim }}_{\infty }\left[-2e^{\frac{-c}{2}}-\left(-2e^{\frac{-b}{2}}\right)\right]\\ ={}_c{\underrightarrow{\lim }}_{\infty }\left[-2e^{\frac{-c}{2}}+2e^{\frac{-b}{2}}\right]\\ =0+2e^{\frac{-b}{2}}\\ =2e^{\frac{-b}{2}}\end{array}$

As $b\to \infty ,2e^{\frac{-b}{2}}\to 0$ , so for large enough b , ${\displaystyle {\int }_b^{\infty }f\left(x\right)dx}$ is as small as one want. Likewise, for large enough b , ${\displaystyle {\int }_{-b}^{\infty }f\left(x\right)dx}$ is as small as a person wants.