Chapter 5.9, Problem 4E

(a)

To determine

To sketch:

The graph of the given function.

Explanation of Solution

Given:

The curve

  $\mathrm{I}={\displaystyle \underset{0}{\overset{1}{\int }}\sin \left(\frac{1}{2}{\mathrm{x}}^2\right)}$

Concept used:

Replace the inequality sign and sketch the graph of the resulting equation (use a dashed line for < or > and a solid line for $\le \mathrm{o}\mathrm{r}\ge$

Test one point in each of the region formed by the graph

If the point satisfies the inequality then shade the entire region to denote that every point in the region satisfies the inequality

Calculation:

The curve

  $\mathrm{I}={\displaystyle \underset{0}{\overset{1}{\int }}\sin \left(\frac{1}{2}{\mathrm{x}}^2\right)}$

Is concave up and increasing in this interval

Draw the graph

  

  ${\mathrm{L}}_2:$ underestimate first rectangle is $0$ and second is missing.

  ${\mathrm{R}}_2:$ overestimate large parts of the rectangle is outside area

  ${\mathrm{M}}_2:$ underestimate(the areas of the rectangle outside the curve are less than the area under the curve is not in rectangle because the rate of increasing is increasing.)

  ${\mathrm{T}}_2:$ overestimate (trapezoids is always overestimate for portions of the curve that are concave up

(b)

To determine

To find:

The vaue of $\mathrm{n}$ of the given function.

Explanation of Solution

Given:

The curve

  $\mathrm{I}={\displaystyle \underset{0}{\overset{1}{\int }}\sin \left(\frac{1}{2}{\mathrm{x}}^2\right)}$

Concept used:

Replace the inequality sign and sketch the graph of the resulting equation (use a dashed line for < or > and a solid line for $\le \mathrm{o}\mathrm{r}\ge$

Test one point in each of the region formed by the graph

If the point satisfies the inequality then shade the entire region to denote that every point in the region satisfies the inequality

Calculation:

The curve

  $\mathrm{I}={\displaystyle \underset{0}{\overset{1}{\int }}\sin \left(\frac{1}{2}{\mathrm{x}}^2\right)}$

Is concave up and increasing in this interval

Draw the graph

  

  ${\mathrm{L}}_2:$ underestimate first rectangle is $0$ and second is missing.

  ${\mathrm{R}}_2:$ overestimate large parts of the rectangle is outside area

  ${\mathrm{M}}_2:$ underestimate(the areas of the rectangle outside the curve are less than the area under the curve is not in rectangle because the rate of increasing is increasing.)

  ${\mathrm{T}}_2:$ overestimate (trapezoids is always overestimate for portions of the curve that are concave up

  ${\mathrm{L}}_{\mathrm{n}}<{\mathrm{M}}_{\mathrm{n}}<\mathrm{I}<{\mathrm{T}}_{\mathrm{n}}<{\mathrm{R}}_{\mathrm{n}}$

  $\mathrm{I}$ is greater than the underestimates.less than the overestimates.

(c)

To determine

To explain:

The vaue of ${\mathrm{L}}_5,{\mathrm{R}}_5,{\mathrm{M}}_5,{\mathrm{T}}_5$ of the given function.

Explanation of Solution

Given:

The curve

  $\mathrm{I}={\displaystyle \underset{0}{\overset{1}{\int }}\sin \left(\frac{1}{2}{\mathrm{x}}^2\right)}$

Concept used:

Formula:

  $\mathrm{\Delta }\mathrm{x}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}$

Calculation:

The curve

  $\mathrm{I}={\displaystyle \underset{0}{\overset{1}{\int }}\sin \left(\frac{1}{2}{\mathrm{x}}^2\right)}$

Intervals $\mathrm{n}=5$

  $\mathrm{\Delta }\mathrm{x}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}=\frac{1}{5}=0.2$

Left end point method at $\mathrm{x}=0$

$\mathrm{x}$	$0.000$	$0.200$	$0.400$	$0.600$	$0.800$
$\sin \left(\frac{1}{2}{\mathrm{x}}^2\right)$	$0.00000$	$0.02000$	$0.07991$	$0.17903$	$0.31457$

  $\begin{array}{l}{\mathrm{L}}_5={\displaystyle \underset{0}{\overset{1}{\int }}\sin \left(\frac{1}{2}{\mathrm{x}}^2\right)}\mathrm{d}\mathrm{x}\\ {\mathrm{L}}_5\approx \mathrm{\Delta }\mathrm{x}\left(0+0.02000+0.07991+0.17903+0.31457\right)\\ {\mathrm{L}}_5\approx 0.2\left(0+0.02000+0.07991+0.17903+0.31457\right)\\ {\mathrm{L}}_5\approx 0.11870\end{array}$

Right end point method at $\mathrm{x}=0.2$

$\mathrm{x}$	$1.000$	$0.200$	$0.400$	$0.600$	$0.800$
$\sin \left(\frac{1}{2}{\mathrm{x}}^2\right)$	$0.47943$	$0.02000$	$0.07991$	$0.17903$	$0.31457$

  $\begin{array}{l}{\mathrm{R}}_5={\displaystyle \underset{0}{\overset{1}{\int }}\sin \left(\frac{1}{2}{\mathrm{x}}^2\right)}\mathrm{d}\mathrm{x}\\ {\mathrm{R}}_5\approx \mathrm{\Delta }\mathrm{x}\left(0.47943+0.02000+0.07991+0.17903+0.31457\right)\\ {\mathrm{R}}_5\approx 0.2\left(0.47943+0.02000+0.07991+0.17903+0.31457\right)\\ {\mathrm{R}}_5\approx 0.21459\end{array}$

Mid point method at $\mathrm{x}=\frac{0.2}{2}=0.1$

$\mathrm{x}$	$0.100$	$0.300$	$0.500$	$0.700$	$0.900$
$\sin \left(\frac{1}{2}{\mathrm{x}}^2\right)$	$0.00500$	$0.04498$	$0.12467$	$0.24256$	$0.39402$

  $\begin{array}{l}{\mathrm{M}}_5={\displaystyle \underset{0}{\overset{1}{\int }}\sin \left(\frac{1}{2}{\mathrm{x}}^2\right)}\mathrm{d}\mathrm{x}\\ {\mathrm{M}}_5\approx \mathrm{\Delta }\mathrm{x}\left(0.00500+0.04498+0.12467+0.24256+0.39402\right)\\ {\mathrm{M}}_5\approx 0.1\left(0.00500+0.04498+0.12467+0.24256+0.39402\right)\\ {\mathrm{M}}_5\approx 0.61225\end{array}$

Trapezoid method at $\mathrm{x}=0$

$\mathrm{x}$	$0.000$	$0.200$	$0.400$	$0.600$	$0.800$	$1.000$
$\sin \left(\frac{1}{2}{\mathrm{x}}^2\right)$	$0.00000$	$0.02000$	$0.07991$	$0.17903$	$0.31457$	$0.47943$

  $\begin{array}{l}{\mathrm{T}}_5={\displaystyle \underset{0}{\overset{1}{\int }}\sin \left(\frac{1}{2}{\mathrm{x}}^2\right)}\mathrm{d}\mathrm{x}\\ {\mathrm{T}}_5\approx \mathrm{\Delta }\mathrm{x}\left(0+0.02000+0.07991+0.17903+0.31457\right)\\ {\mathrm{T}}_5\approx 0.1\left(0+2\left(0.02000\right)+2\left(0.07991\right)+2\left(0.17903\right)+2\left(0.31457\right)+0.47943\right)\\ {\mathrm{T}}_5\approx 0.16665\end{array}$