Trapezoidal Rule

Question 1 of 5

Approximate the area under the curve

$y=10/x$ using Trapezoidal rule.

$text(Area) =$ (16.8333) $\text(square units)$

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The Trapezoidal Rule approximates the area under a curve by dividing it into several trapezia.

$A ≈ h/2 [y_0 + y_L +2(y_1 + y_2 + y_3 + … + y_(L-1)]$

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

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

$=$

$(10)/x$

$a=1$ (lower limit)

$b=5$ (upper limit)

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

Solve for

$h$

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

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

	$=$	$4/4$	Simplify

	$=$	$1$

Construct a table of values to find the

$\text(y-values)$ for each

$\text(x-value)$

$y=10/x$

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

Substitute $x=1$ into the given equation.

$y$	$=$	$\frac {10}{\color{#9a00c7}{1}}$
$y_0$	$=$	$10$

$x$	$1$	$2$	$3$	$4$	$5$
$y$	$10$

Substitute $x=2$ into the given equation.

$y$	$=$	$\frac {10}{\color{#9a00c7}{2}}$
$y_1$	$=$	$5$

$x$	$1$	$2$	$3$	$4$	$5$
$y$	$10$	$5$

Repeat this process for each

$\text(x-value)$

$x$	$1$	$2$	$3$	$4$	$5$
$y$	$10$	$5$	$10/3$	$10/4$	$2$

Apply the Trapezoidal Rule

$A$	$≈$	$h/2 [$ $y_0$ $+$ $y_L$ $+ 2($ $y_1$ $+$ $y_2$ $+$ $y_3$ $+ … +$ $y_(L-1)$ $]$	Trapezoidal Rule formula

	$≈$	$1/2 [$ $10$ $+$ $2$ $+2($ $5$ $+$ $(10)/3$ $+$ $(10)/4)$ $]$	$h=1$

	$≈$	$1/2 [12+2((65)/6)]$	Simplify

	$≈$	$1/2 [(101)/3]$

	$≈$	$16.8333$

$16.8333 \text(square units)$

Question 2 of 5

Approximate the area under the curve

$y=1/(2x-1)$ using Trapezoidal rule.

$text(Area) =$ (0.39554) $\text(square units)$

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The Trapezoidal Rule approximates the area under a curve by dividing it into several trapezia.

Trapezoidal Rule

$A ≈ h/2 [y_0 + y_L +2(y_1 + y_2 + y_3 + … + y_(L-1)]$

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

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

$=$

$1/(2x-1)$

$a=3$ (lower limit)

$b=6$ (upper limit)

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

Solve for

$h$

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

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

	$=$	$3/6$	Simplify

	$=$	$1/2$

Construct a table of values to find the

$\text(y-values)$ for each

$\text(x-value)$

$y=1/(2x-1)$

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

Substitute $x=3$ into the given equation.

$y$	$=$	$\frac {1}{2 \left(\color{#9a00c7}{3} \right) -1}$

	$=$	$1/(6-1)$

$y_0$	$=$	$1/5$

$x$	$3$	$3.5$	$4$	$4.5$	$5$	$5.5$	$6$
$y$	$1/5$

Substitute $x=3.5$ into the given equation.

$y$	$=$	$\frac {1}{2 \left(\color{#9a00c7}{3.5} \right) -1}$

	$=$	$1/(7-1)$

$y_1$	$=$	$1/6$

$x$	$3$	$3.5$	$4$	$4.5$	$5$	$5.5$	$6$
$y$	$1/5$	$1/6$

Repeat this process for each

$\text(x-value)$

$x$	$3$	$3.5$	$4$	$4.5$	$5$	$5.5$	$6$
$y$	$1/5$	$1/6$	$1/7$	$1/8$	$1/9$	$1/10$	$1/11$

Apply the Trapezoidal Rule

$A$	$≈$	$h/2 [$ $y_0$ $+$ $y_L$ $+ 2($ $y_1$ $+$ $y_2$ $+$ $y_3$ $+ … +$ $y_(L-1)$ $]$	Trapezoidal Rule formula

	$≈$	$(1/2)/2 [$ $1/5$ $+$ $1/11$ $+2($ $1/6$ $+$ $1/7$ $+$ $1/8$ $+$ $1/9$ $+$ $1/10$ $+$ $1/11$ $)]$	$h=1/2$

	$≈$	$1/4 [16/55 +2((1627)/(2520))]$	Simplify

	$≈$	$1/4 [1.582179]$

	$≈$	$0.39554$

$0.39554 \text(square units)$

Question 3 of 5

Approximate the area under the curve

$y=4^x$ using Trapezoidal rule.

$text(Area) =$ (2.1855) $\text(square units)$

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The Trapezoidal Rule approximates the area under a curve by dividing it into several trapezia.

Trapezoidal Rule

$A ≈ h/2 [y_0 + y_L +2(y_1 + y_2 + y_3 + … + y_(L-1)]$

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

$\int_{\color{#9a00c7}{0}}^{\color{#007ddc}{1}} y$

$=$

$4^x$

$a=0$ (lower limit)

$b=1$ (upper limit)

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

Solve for

$h$

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

	$=$	$\frac {\color {#007ddc}{1}-\color {#9a00c7}{0}}{\color {#00880a}{4}}$	Substitute values of $a$ , $b$ , and $n$

	$=$	$1/4$	Simplify

Construct a table of values to find the

$\text(y-values)$ for each

$\text(x-value)$

$y=4^x$

$x$	$0$	$1/4$	$1/2$	$3/4$	$1$
$y$

Substitute $x=0$ into the given equation.

$y$	$=$	$4^{\color{#9a00c7}{0}}$
$y_0$	$=$	$1$

$x$	$0$	$1/4$	$1/2$	$3/4$	$1$
$y$	$1$

Substitute $x=1/4$ into the given equation.

$y$	$=$	$4^{\color{#9a00c7}{1/4}}$
$y_1$	$=$	$4^(1/4)$

$x$	$0$	$1/4$	$1/2$	$3/4$	$1$
$y$	$1$	$4^(1/4)$

Repeat this process for each

$\text(x-value)$

$x$	$0$	$1/4$	$1/2$	$3/4$	$1$
$y$	$1$	$4^(1/4)$	$2$	$4^(3/4)$	$4$

Apply the Trapezoidal Rule

$A$	$≈$	$h/2 [$ $y_0$ $+$ $y_L$ $+ 2($ $y_1$ $+$ $y_2$ $+$ $y_3$ $+ … +$ $y_(L-1)$ $]$	Trapezoidal Rule formula

	$≈$	$(1/4)/2 [$ $1$ $+$ $4$ $+2($ $4^(1/4)$ $+$ $2$ $+$ $4^(3/4)$ $]$	$h=1/4$

	$≈$	$1/8 [5+2(6.242)]$	Simplify

	$≈$	$1/8 [5+12.484]$

	$≈$	$1/8 [17.484]$

	$≈$	$2.1855$

$2.1855 \text(square units)$

Question 4 of 5

Approximate the area under the curve

$y=(2x)/(1+x)$ using Trapezoidal rule.

$text(Area) =$ (139/30) $\text(square units)$

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The Trapezoidal Rule approximates the area under a curve by dividing it into several trapezia.

Trapezoidal Rule

$A ≈ h/2 [y_0 + y_L +2(y_1 + y_2 + y_3 + … + y_(L-1)]$

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

$\int_{\color{#9a00c7}{0}}^{\color{#007ddc}{4}} y$

$=$

$(2x)/(1+x)$

$a=0$ (lower limit)

$b=4$ (upper limit)

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

Solve for

$h$

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

	$=$	$\frac {\color {#007ddc}{4}-\color {#9a00c7}{0}}{\color {#00880a}{4}}$	Substitute values of $a$ , $b$ , and $n$

	$=$	$4/4$	Simplify

	$=$	$1$

Construct a table of values to find the

$\text(y-values)$ for each

$\text(x-value)$

$y=(2x)/(1+x)$

$x$	$0$	$1$	$2$	$3$	$4$
$y$

Substitute $x=0$ into the given equation.

$y$	$=$	$\frac {2 \left(\color{#9a00c7}{0}\right)}{1+\color{#9a00c7}{0}}$

	$=$	$0/1$

$y_0$	$=$	$0$

$x$	$0$	$1$	$2$	$3$	$4$
$y$	$0$

Substitute $x=1$ into the given equation.

$y$	$=$	$\frac {2\left( \color{#9a00c7}{1}\right)}{1+\color{#9a00c7}{1}}$

	$=$	$2/2$

$y_1$	$=$	$1$

$x$	$0$	$1$	$2$	$3$	$4$
$y$	$0$	$1$

Repeat this process for each

$\text(x-value)$

$x$	$0$	$1$	$2$	$3$	$4$
$y$	$0$	$1$	$4/3$	$6/4$	$8/5$

Apply the Trapezoidal Rule

$A$	$≈$	$h/2 [$ $y_0$ $+$ $y_L$ $+ 2($ $y_1$ $+$ $y_2$ $+$ $y_3$ $+ … +$ $y_(L-1)$ $]$	Trapezoidal Rule formula

	$≈$	$1/2 [$ $0$ $+$ $8/5$ $+2($ $1$ $+$ $4/3$ $+$ $6/4$ $)]$	$h=1$

	$≈$	$1/2 [8/5+2((23)/6)]$	Simplify

	$≈$	$1/2 [(139)/15]$

	$≈$	$139/30$

$139/30 \text(square units)$

Question 5 of 5

Find the entire surface area of the small lake.

$text(Area) =$ (1560) $\text(square units)$

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The Trapezoidal Rule approximates the area under a curve by dividing it into several trapezia.

Trapezoidal Rule

$A ≈ h/2 [y_0 + y_L +2(y_1 + y_2 + y_3 + … + y_(L-1)]$

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

$a=0$ (lower limit)

$b=50$ (upper limit)

$n=5$ (number of strips in given diagram)

Solve for

$h$

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

	$=$	$\frac {\color {#007ddc}{50}-\color {#9a00c7}{0}}{\color {#00880a}{5}}$	Substitute values of $a$ , $b$ , and $n$

	$=$	$50/5$	Simplify

	$=$	$10$

Construct a table of values to find the

$\text(y-values)$ for each

$\text(x-value)$ based on the given illustration.

$x$	$0$	$10$	$20$	$30$	$40$	$50$
$y$	$0$	$35$	$32$	$43$	$46$	$0$

Apply the Trapezoidal Rule

$A$	$≈$	$h/2 [$ $y_0$ $+$ $y_L$ $+ 2($ $y_1$ $+$ $y_2$ $+$ $y_3$ $+ … +$ $y_(L-1)$ $]$	Trapezoidal Rule formula

	$≈$	$10/2 [$ $0$ $+$ $0$ $+2($ $35$ $+$ $32$ $+$ $43$ $+$ $46$ $)]$	$h=10$

	$≈$	$5 [2(156)]$	Simplify

	$≈$	$5 [312]$

	$≈$	$1560$

$1560 \text(square units)$

Trapezoidal Rule

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