Chapter 8.4, Problem 57E

(a)

To determine

To show: ${\displaystyle \underset{1}{\overset{\infty }{\int }}{e}^{-x^2}}dx\le {\displaystyle \underset{6}{\overset{\infty }{\int }}{e}^{-6x}dx}<4\times {10}^{-17}$

Answer to Problem 57E

The given expression is shown.

Explanation of Solution

Given:

  ${\displaystyle \underset{1}{\overset{\infty }{\int }}{e}^{-x^2}}dx$

Calculation:

  ${\displaystyle \underset{1}{\overset{\infty }{\int }}{e}^{-x^2}}dx\le {\displaystyle \underset{6}{\overset{\infty }{\int }}{e}^{-6x}dx}<4\times {10}^{-17}$

  $\begin{array}{l}\text{Let }f(x)=e^{-x^2};g\left(x\right)=e^{-6x}\\ x\ge 6;x^2\ge 6x\end{array}$

Hence,

  $e^{-x^2}\le e^{-6x}$

Therefore,

  ${\displaystyle \underset{1}{\overset{\infty }{\int }}{e}^{-x^2}}dx\le {\displaystyle \underset{6}{\overset{\infty }{\int }}{e}^{-6x}dx}$

First part of the given expression is proved.

Now,

  $\begin{array}{l}{\displaystyle \underset{6}{\overset{\infty }{\int }}{e}^{-6x}dx}=\underset{b\to \infty }{\lim }{\displaystyle \underset{6}{\overset{b}{\int }}{e}^{-6x}dx}\\ \text{Let }u=-6x\\ du=-6dx\\ -\frac{1}{6}du=dx\\ {\displaystyle \underset{6}{\overset{\infty }{\int }}{e}^{-6x}dx}=\underset{b\to \infty }{\lim }{\displaystyle \underset{6}{\overset{b}{\int }}-\frac{1}{6}{e}^{u}du}\\ \text{ = }-\frac{1}{6}\underset{b\to \infty }{\lim }\left[e^{-6b}-e^{-36}\right]\\ \text{ = }\frac{1}{6}{e}^{-36}\\ \text{ = 3}\text{.86}\times {\text{10}}^{-17}\end{array}$

As, $\text{ 3}\text{.86}\times {\text{10}}^{-17}\le 4\times {10}^{-17}$

Therefore, ${\displaystyle \underset{1}{\overset{\infty }{\int }}{e}^{-x^2}}dx\le {\displaystyle \underset{6}{\overset{\infty }{\int }}{e}^{-6x}dx}<4\times {10}^{-17}$ is shown.

Conclusion:

The given expression is shown.

(b)

To determine

To explain: ${\displaystyle \underset{1}{\overset{\infty }{\int }}{e}^{-x^2}}dx\approx {\displaystyle \underset{1}{\overset{6}{\int }}{e}^{-x^2}}dx$ with an error of $4\times {10}^{-17}$

Answer to Problem 57E

The given expression is explained.

Explanation of Solution

Given:

  $f\left(x\right)$ and $g\left(x\right)$ becomes and remains negative as $x\to \infty$

Calculation:

  ${\displaystyle \underset{1}{\overset{6}{\int }}{e}^{-x^2}dx+{\displaystyle \underset{6}{\overset{\infty }{\int }}{e}^{-x^2}\le {\displaystyle \underset{1}{\overset{6}{\int }}{e}^{-x^2}dx+4\times {10}^{-17}}}}$

This need to be proved.

As, we know:

  $\begin{array}{l}{\displaystyle \underset{1}{\overset{6}{\int }}{e}^{-x^2}dx+{\displaystyle \underset{6}{\overset{\infty }{\int }}{e}^{-x^2}\le {\displaystyle \underset{1}{\overset{6}{\int }}{e}^{-x^2}dx+4\times {10}^{-17}}}}\\ {\displaystyle \underset{1}{\overset{\infty }{\int }}{e}^{-x^2}}={\displaystyle \underset{1}{\overset{6}{\int }}{e}^{-x^2}dx}+{\displaystyle \underset{6}{\overset{\infty }{\int }}{e}^{-x^2}}\\ \text{As, }{\displaystyle \underset{6}{\overset{\infty }{\int }}{e}^{-x^2}\le 4\times {10}^{-17}}(\text{from(a))}\\ \therefore {\displaystyle \underset{1}{\overset{\infty }{\int }}{e}^{-x^2}dx}\approx {\displaystyle \underset{1}{\overset{6}{\int }}{e}^{-x^2}}dx\end{array}$

Conclusion:

The given expression is explained.

(c)

To determine

Estimate the value of ${\displaystyle \underset{1}{\overset{\infty }{\int }}{e}^{-x^2}dx}$

Answer to Problem 57E

The value of ${\displaystyle \underset{1}{\overset{\infty }{\int }}{e}^{-x^2}dx}$ is $0.13940$

Explanation of Solution

Given:

  ${\displaystyle \underset{1}{\overset{\infty }{\int }}{e}^{-x^2}}dx$

Calculation:

  $\begin{array}{l}{\displaystyle \underset{1}{\overset{\infty }{\int }}{e}^{-x^2}dx=\underset{b\to \infty }{\lim }}{\displaystyle \underset{1}{\overset{b}{\int }}{e}^{-x^2}dx}\\ {\displaystyle \int e^{-x^2}dx}=\frac{\sqrt{\pi }erf\left(x\right)}{2}\\ \underset{b\to \infty }{\lim }{\displaystyle \underset{1}{\overset{b}{\int }}{e}^{-x^2}dx}=\underset{b\to \infty }{\lim }\frac{\sqrt{\pi }erf\left(b\right)}{2}-\frac{\sqrt{\pi }erf\left(1\right)}{2}\\ \underset{b\to \infty }{\lim }{\displaystyle \underset{1}{\overset{b}{\int }}{e}^{-x^2}dx}=\frac{\sqrt{\pi }}{2}-\frac{\sqrt{\pi }erf\left(1\right)}{2}\\ \underset{b\to \infty }{\lim }{\displaystyle \underset{1}{\overset{b}{\int }}{e}^{-x^2}dx}\approx 0.13940\end{array}$

Conclusion:

The value of ${\displaystyle \underset{1}{\overset{\infty }{\int }}{e}^{-x^2}dx}$ is $0.13940$

(d)

To determine

To explain: ${\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-x^2}dx}={\displaystyle \underset{0}{\overset{3}{\int }}{e}^{-x^2}dx}$ with an error of $0.000042$

Answer to Problem 57E

The given expression is explained.

Explanation of Solution

Given:

  ${\displaystyle \underset{1}{\overset{\infty }{\int }}{e}^{-x^2}}dx$

Calculation:

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-x^2}dx}={\displaystyle \underset{0}{\overset{3}{\int }}{e}^{-x^2}dx}+{\displaystyle \underset{3}{\overset{\infty }{\int }}{e}^{-x^2}dx}\le {\displaystyle \underset{0}{\overset{3}{\int }}{e}^{-x^2}dx+{\displaystyle \underset{3}{\overset{\infty }{\int }}{e}^{-3x}dx}}\\ As,x^2\ge 3x\text{ for all x > 3}\\ {\text{Then, e}}^{-x^2}\le e^{-3x}\end{array}$

Now, evaluate the integral

  $\begin{array}{l}{\displaystyle \underset{3}{\overset{\infty }{\int }}{e}^{-3x}}dx=\underset{b\to \infty }{\lim }{\displaystyle \underset{3}{\overset{b}{\int }}{e}^{-3x}}dx\\ \text{Let }u=-3x\\ du=-3dx\\ \text{ -}\frac{1}{3}du=dx\\ {\displaystyle \underset{3}{\overset{\infty }{\int }}{e}^{-3x}}dx=\underset{b\to \infty }{\lim }{\displaystyle \underset{3}{\overset{b}{\int }}-\frac{1}{3}{e}^u}du\\ \text{ = -}\frac{1}{3}\underset{b\to \infty }{\lim }\left[e^{-3b}-e^{-9}\right]\\ \text{ = }\frac{1}{3}{e}^{-9}\\ \text{ =0}\text{.000041}\\ \text{0}\text{.000041}<00.000042\end{array}$

Hence, proved.

Conclusion:

The given expression is explained.