Trapezoidal Rule
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Question 1 of 5
1. Question
Approximate the area under the curve `y=10/x` using Trapezoidal rule.
- `text(Area) =` (16.8333) `\text(square units)`
Hint
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Well Done!
Incorrect
Remember
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$$\int_{\color{#9a00c7}{1}}^{\color{#007ddc}{5}} y$$ `=` `(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
2. Question
Approximate the area under the curve `y=1/(2x-1)` using Trapezoidal rule.
- `text(Area) =` (0.39554) `\text(square units)`
Hint
Help VideoCorrect
Nice Job!
Incorrect
Remember
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
3. Question
Approximate the area under the curve `y=4^x` using Trapezoidal rule.
- `text(Area) =` (2.1855) `\text(square units)`
Hint
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Excellent!
Incorrect
Remember
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
4. Question
Approximate the area under the curve `y=(2x)/(1+x)` using Trapezoidal rule.
- `text(Area) =` (139/30) `\text(square units)`
Hint
Help VideoCorrect
Fantastic!
Incorrect
Remember
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
5. Question
Find the entire surface area of the small lake.
- `text(Area) =` (1560) `\text(square units)`
Hint
Help VideoCorrect
Keep Going!
Incorrect
Remember
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)`
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