Simpsons Rule

Question 1 of 7

Find the area under the curve

y = \frac{x^{2}}{4}

$y=x^2/4$

$Area =$ $\text(Area) =$ (4.6667) $square units$ $\text(square units)$

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\int_{a}^{b} f (x) d x \approx \frac{b - a}{6} [f (a) + 4 f (\frac{a + b}{2}) + f (b)]

$int_a^b f(x) dx ≈ (b-a)/6 [f(a)+4f((a+b)/2)+f(b)]$

First, construct a table of values for both

x

$x$ and

y

$y$ given the equation.

y = \frac{x^{2}}{4}

$y=x^2/4$

$x$ $x$	$2$ $2$	$3$ $3$	$4$ $4$
$y$ $y$

Substitute $x = 2$ $x=2$ into the given equation.

$y$ $y$	$=$ $=$	$\frac {\color{#9932CC}{2}^2}{4}$	Substitute the value of $x$ $x$
	$=$ $=$	$\frac{4}{4}$ $4/4$	Simplify
$y$ $y$	$=$ $=$	$1$ $1$
$f (a)$ $f(a)$	$=$ $=$	$1$ $1$

$x$ $x$	$2$ $2$	$3$ $3$	$4$ $4$
$y$ $y$	$1$ $1$

Substitute $x = 3$ $x=3$ into the given equation.

$y$ $y$	$=$ $=$	$\frac {\color{#9932CC}{3}^2}{4}$	Substitute the value of $x$ $x$
$y$ $y$	$=$ $=$	$\frac{9}{4}$ $9/4$	Simplify
$f (\frac{a + b}{2})$ $f((a+b)/2)$	$=$ $=$	$\frac{9}{4}$ $9/4$

$x$ $x$	$2$ $2$	$3$ $3$	$4$ $4$
$y$ $y$	$1$ $1$	$\frac{9}{4}$ $9/4$

Repeat this process for each

x-value

$\text(x-value)$

$x$ $x$	$2$ $2$	$3$ $3$	$4$ $4$
$y$ $y$	$1$ $1$	$\frac{9}{4}$ $9/4$	$4$ $4$

Apply Simpsons Rule

$A$ $A$	$\approx$ $≈$	$\int_{\color {#9a00c7}{a}}^{\color{#007ddc}{b}} f(x) dx$

	$\approx$ $≈$	$\frac{\color{#007ddc}{b}- \color{#9a00c7}{a}}{6} [\color {#00880a}{f(a)} +4 \color{#00880a}{\frac{f(a+b)}{2}}+\color {#00880a}{f(b)}]$	Simpsons Rule formula

	$\approx$ $≈$	$\frac{\color{#007ddc}{4}- \color{#9a00c7}{2}}{6} [\color {#00880a}{1} +4 \color{#00880a}{(\frac{9}{4})}+\color {#00880a}{4}]$	Substitute known values

	$\approx$ $≈$	$\frac{2}{6} [5 + \frac{36}{4}]$ $2/6 [5+36/4]$	Simplify

	$\approx$ $≈$	$\frac{1}{3} [14]$ $1/3 [14]$

	$\approx$ $≈$	$\frac{14}{3}$ $14/3$

	$\approx$ $≈$	$4.6667$ $4.6667$

4.6667 square units

$4.6667 \text(square units)$

Question 2 of 7

Find the area under the curve

y = 2^{x}

$y=2^x$

$Area =$ $\text(Area) =$ (43.333)

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Simpsons Rule

\int_{a}^{b} f (x) d x \approx \frac{h}{3} [y_{0} + y_{L} + 2 (y_{2} + y_{4} + \dots) + 4 (y_{1} + y_{3} + \dots)]

$int_a^b f(x) dx ≈ h/3 [y_0+y_L+2(y_2+y_4+…)+4(y_1+y_3+…)]$

First, find the values of $a$ $a$ , $b$ $b$ and $n$ $n$ from the given equation

\int_{1}^{5} y

$\int_{\color{#9a00c7}{1}}^{\color{#007ddc}{5}} y$

=

$=$

2^{x}

$2^x$

$a = 1$ $a=1$ (lower limit)

$b = 5$ $b=5$ (upper limit)

$n = 4$ $n=4$ (number of strips in given diagram)

Solve for

h

$h$

$h$ $h$	$=$ $=$	$\frac {\color {#007ddc}{b}-\color {#9a00c7}{a}}{\color {#00880a}{n}}$

	$=$ $=$	$\frac {\color {#007ddc}{5}-\color {#9a00c7}{1}}{\color {#00880a}{4}}$	Substitute values of $a$ $a$ , $b$ $b$ , and $n$ $n$

	$=$ $=$	$\frac{4}{4}$ $4/4$	Simplify

$h$ $h$	$=$ $=$	$1$ $1$

First, construct a table of values for both

x

$x$ and

y

$y$ given the equation.

y = 2^{x}

$y=2^x$

$x$ $x$	$1$ $1$	$2$ $2$	$3$ $3$	$4$ $4$	$5$ $5$
$y$ $y$

Substitute $x = 1$ $x=1$ into the given equation.

$y$ $y$	$=$ $=$	$2^{\color{#9932CC}{1}}$	Substitute the value of $x$ $x$
	$=$ $=$	$2^{1}$ $2^1$	Simplify
$y_{0}$ $y_0$	$=$ $=$	$2$ $2$

$x$ $x$	$1$ $1$	$2$ $2$	$3$ $3$	$4$ $4$	$5$ $5$
$y$ $y$	$2$ $2$

Substitute $x = 2$ $x=2$ into the given equation.

$y$ $y$	$=$ $=$	$2^{\color{#9932CC}{2}}$	Substitute the value of $x$ $x$
$y_{1}$ $y_1$	$=$ $=$	$4$ $4$	Simplify

$x$ $x$	$1$ $1$	$2$ $2$	$3$ $3$	$4$ $4$	$5$ $5$
$y$ $y$	$2$ $2$	$4$ $4$

Repeat this process for each

x-value

$\text(x-value)$

$x$ $x$	$1$ $1$	$2$ $2$	$3$ $3$	$4$ $4$	$5$ $5$
$y$ $y$	$2$ $2$	$4$ $4$	$8$ $8$	$16$ $16$	$32$ $32$

Apply Simpsons Rule

$A$ $A$	$\approx$ $≈$	$\int_{\color {#9a00c7}{a}}^{\color{#007ddc}{b}} f(x) dx$

	$\approx$ $≈$	$\frac{h}{3} [$ $h/3 [$ $y_{0}$ $y_0$ $+$ $+$ $y_{L}$ $y_L$ $+ 2 ($ $+2($ $y_{2}$ $y_2$ $+$ $+$ $y_{4}$ $y_4$ $+ \dots) + 4 ($ $+…)+4($ $y_{1}$ $y_1$ $+$ $+$ $y_{3}$ $y_3$ $+ \dots)]$ $+…)]$	Simpsons Rule formula

	$\approx$ $≈$	$\frac{1}{3} [$ $1/3 [$ $2$ $2$ $+$ $+$ $32$ $32$ $+ 2 ($ $+2($ $8$ $8$ $) + 4 ($ $)+4($ $4$ $4$ $+$ $+$ $16$ $16$ $)]$ $)]$	Substitute known values

	$\approx$ $≈$	$\frac{1}{3} [34 + 16 + 80]$ $1/3 [34+16+80]$	Simplify

	$\approx$ $≈$	$\frac{1}{3} [130]$ $1/3 [130]$

	$\approx$ $≈$	$\frac{130}{3}$ $(130)/3$

	$\approx$ $≈$	$43.333$ $43.333$

43.333

$43.333$

Question 3 of 7

Find the area under the curve

y = \frac{6}{x}

$y=6/x$

$Area =$ $\text(Area) =$ (4.159)

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Simpsons Rule

\int_{a}^{b} f (x) d x \approx \frac{h}{3} [y_{0} + y_{L} + 2 (y_{2} + y_{4} + \dots) + 4 (y_{1} + y_{3} + \dots)]

$int_a^b f(x) dx ≈ h/3 [y_0+y_L+2(y_2+y_4+…)+4(y_1+y_3+…)]$

First, find the values of $a$ $a$ , $b$ $b$ and $n$ $n$ from the given equation

\int_{3}^{6} y

$\int_{\color{#9a00c7}{3}}^{\color{#007ddc}{6}} y$

=

$=$

\frac{6}{x}

$6/x$

$a = 3$ $a=3$ (lower limit)

$b = 6$ $b=6$ (upper limit)

$n = 6$ $n=6$ (number of strips in given diagram)

Solve for

h

$h$

$h$ $h$	$=$ $=$	$\frac {\color {#007ddc}{b}-\color {#9a00c7}{a}}{\color {#00880a}{n}}$

	$=$ $=$	$\frac {\color {#007ddc}{6}-\color {#9a00c7}{3}}{\color {#00880a}{6}}$	Substitute values of $a$ $a$ , $b$ $b$ , and $n$ $n$

	$=$ $=$	$\frac{3}{6}$ $3/6$	Simplify

$h$ $h$	$=$ $=$	$\frac{1}{2}$ $1/2$

First, construct a table of values for both

x

$x$ and

y

$y$ given the equation.

y = 2^{x}

$y=2^x$

$x$ $x$	$3$ $3$	$3.5$ $3.5$	$4$ $4$	$4.5$ $4.5$	$5$ $5$	$5.5$ $5.5$	$6$ $6$
$y$ $y$

Substitute $x = 3$ $x=3$ into the given equation.

$y$ $y$	$=$ $=$	$\frac {6}{\color{#9932CC}{3}}$	Substitute the value of $x$ $x$
	$=$ $=$	$2$ $2$	Simplify
$y_{0}$ $y_0$	$=$ $=$	$2$ $2$

$x$ $x$	$3$ $3$	$3.5$ $3.5$	$4$ $4$	$4.5$ $4.5$	$5$ $5$	$5.5$ $5.5$	$6$ $6$
$y$ $y$	$2$ $2$

Substitute $x = 3.5$ $x=3.5$ into the given equation.

$y$ $y$	$=$ $=$	$\frac{6}{\color{#9932CC}{3.5}}$	Substitute the value of $x$ $x$
$y_{1}$ $y_1$	$=$ $=$	$\frac{6}{3.5}$ $6/3.5$	Simplify

$x$ $x$	$3$ $3$	$3.5$ $3.5$	$4$ $4$	$4.5$ $4.5$	$5$ $5$	$5.5$ $5.5$	$6$ $6$
$y$ $y$	$2$ $2$	$\frac{6}{3.5}$ $6/3.5$

Repeat this process for each

x-value

$\text(x-value)$

$x$ $x$	$3$ $3$	$3.5$ $3.5$	$4$ $4$	$4.5$ $4.5$	$5$ $5$	$5.5$ $5.5$	$6$ $6$
$y$ $y$	$2$ $2$	$\frac{6}{3.5}$ $6/3.5$	$\frac{6}{4}$ $6/4$	$\frac{6}{4.5}$ $6/4.5$	$\frac{6}{5}$ $6/5$	$\frac{6}{5.5}$ $6/5.5$	$1$ $1$

Apply Simpsons Rule

$A$ $A$	$\approx$ $≈$	$\int_{\color {#9a00c7}{a}}^{\color{#007ddc}{b}} f(x) dx$

	$\approx$ $≈$	$\frac{h}{3} [$ $h/3 [$ $y_{0}$ $y_0$ $+$ $+$ $y_{L}$ $y_L$ $+ 2 ($ $+2($ $y_{2}$ $y_2$ $+$ $+$ $y_{4}$ $y_4$ $+ \dots) + 4 ($ $+…)+4($ $y_{1}$ $y_1$ $+$ $+$ $y_{3}$ $y_3$ $+ \dots)]$ $+…)]$	Simpsons Rule formula

	$\approx$ $≈$	$\frac{\frac{1}{2}}{3} [$ $(1/2)/3 [$ $2$ $+$ $1$ $+2($ $6/4$ $+$ $6/5$ $)+4($ $6/3.5$ $+$ $6/4.5$ $+$ $6/5.5$ $)]$	Substitute known values

	$≈$	$1/6 [3+2(27/10)+4(4.1385)]$	Simplify

	$≈$	$1/6 [3+54/10+16.554113]$

	$≈$	$1/6 (24.954113)$

	$≈$	$4.159$

$4.159$

Question 4 of 7

Find and estimate the area of the orange parcel of land.

$\text(Area) =$ (1960) $m^2$

Question 5 of 7

Find and estimate the area of the huge block of land.

$\text(Area) =$ (251333 1/3, 251333.33, 251333.3333) $m^2$

Question 6 of 7

Find and estimate the area of the yellow parcel of land.

$\text(Area) =$ (45000) $m^2$

Question 7 of 7

Find and estimate the area of the irregular piece of land.

$\text(Area) =$ (5558 2/3) $m^2$

$y_0$	$y_1$	$y_2$	$y_3$	$y_4$	$y_5$	$y_L$
$0$	$150$	$290$	$320$	$210$	$180$	$170$

$d_f$	$d_m$	$d_L$
$0$	$49$	$68$

$d_f$	$d_m$	$d_L$
$68$	$88$	$74$

Simpsons Rule

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Quiz summary

Information

1. Question

Hint

Great Work!

Simpsons Rule

2. Question

Hint

Correct!

Simpsons Rule

3. Question

Hint

Keep Going!

Simpsons Rule

4. Question

Hint

Fantastic!

Simpsons Rule

5. Question

Hint

Excellent!

Simpsons Rule

6. Question

Hint

Nice Job!

Simpsons Rule

7. Question

Hint

Well Done!

Simpsons Rule

Quizzes

$A$	$≈$	$h/3 [$ $y_0$ $+$ $y_L$ $+2($ $y_2$ $+$ $y_4$ $+…)+4($ $y_1$ $+$ $y_3$ $+…)]$	Simpsons Rule formula

	$≈$	$(20)/3 [$ $21.5$ $+$ $19.5$ $+2($ $22.5$ $)+4($ $24$ $+$ $28$ $)]$	Substitute known values

	$≈$	$20/3 [41+45+208]$	Simplify

	$≈$	$20/3 [294]$

	$≈$	$1960$

$A_1$	$≈$	$h/3 [d_f+4d_m+d_L]$	Simpsons Rule formula

	$≈$	$(22)/3 [$ $0$ $+4($ $49$ $)+$ $68$ $]$	Substitute known values

	$≈$	$22/3 [68+196]$	Simplify

	$≈$	$22/3 [264]$

$A_1$	$≈$	$1936$ $m^2$

$A_2$	$≈$	$h/3 [d_f+4d_m+d_L]$	Simpsons Rule formula

	$≈$	$(22)/3 [$ $68$ $+4($ $88$ $)+$ $74$ $]$	Substitute known values

	$≈$	$22/3 [142+352]$	Simplify

	$≈$	$22/3 [494]$

$A_2$	$≈$	$3622 2/3$ $m^2$

$A$	$≈$	$A_1$ $+$ $A_2$	Combine the areas

	$≈$	$1936$ $+$ $3622 2/3$

	$≈$	$5558 2/3$

$y_0$	$y_1$	$y_2$	$y_3$	$y_L$
$21.5$	$24$	$22.5$	$28$	$19.5$

$y_0$	$y_1$	$y_2$	$y_3$	$y_L$
$110$	$98$	$102$	$124$	$148$