Area of a trapezoid by integration

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Question statement

{applet}

Give any introductory information the student needs.

			Name	Type	Generated Value

A	vector	vector(1.6,4.2)
B	vector	vector(5.6,2)
applet	ggbapplet	HTML node
strings	list	[ "x_1", "y_1", "x_2", "y_2" ]
x1	expression	x_1
y1	expression	y_1
x2	expression	x_2
y2	expression	y_2
area	expression	(x_2 - x_1)(y_2 + y_1)/2
AreaNum	number	12.4

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Name

Data type

Value

xxxxxxxxxx
 
vector(random(-20..30)/10, random(0..50)/10)

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Description

x = 0 is never selected, but problem will still work. Equations in solution will look poor.

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Generated value: `vector`

vector(1.6,4.2)

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AreaNum
B
applet
strings

Variable testing condition

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1. Gap Area Mathematical expression
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Use integration to find an expression for the area under the line from $x = \var{x1}$ to $x = \var{x2}$ .

$A =$ Area

Advice

The equation of a line passing through two points $A = (\var{x1}, \var{y1}) $ and $B = (\var{x2}, \var{y2})$ is

$\dfrac{y-\var{y1}}{\simplify[zeroTerm]{x - {x1}}} = \dfrac{\var{y2}-\var{y1}}{\simplify[zeroTerm]{{x2} - {x1}}}$

which rearranges to:

$y = \left(\dfrac{\var{y2}-\var{y1}}{\simplify[zeroTerm]{{x2} - {x1}}}\right) (\simplify[zeroTerm]{(x - {x1})}) + \var{y1}$

Integrate using vertical strips $dA = y dx$ to find the area

$A = \int dA = \simplify{defint(y, x, {x1},{x2})}$

$\phantom{A} = \int_\var{x1}^\var{x2} \left(\dfrac{\var{y2}-\var{y1}}{\simplify[zeroTerm]{{x2} - {x1}}}\right) (\simplify[zeroTerm]{(x - {x1})}) + \var{y1}\; dx$

Simplify by letting $ m = \frac{\var{y2}-\var{y1}}{\simplify[zeroTerm]{{x2}-{x1}}}$ and performing a $u$-substitiution with

$u = (x - \var{x1}) $ and $du = dx$

and limits $ u(\var{x1} )= 0$, and $u(\var{x2}) = \var{x2}-\var{x1}$ to give:

$A = \int_0^\simplify{{x2}-{x1}} \simplify{(m u + {y1})} \;du$

This function can be easily integrated with respect to $u$, and then expressed in terms of $x$:

$A = m \dfrac{u^2}{2} + \var{y1}\; u \;\Big|_0^\simplify{{x2}-{x1}}$

$\phantom{A} = \dfrac{(\var{y2}-\var{y1})}{(\simplify[zeroTerm]{{x2}-{x1}})} \dfrac{\simplify{({x2}-{x1})^2}}{2} + \var{y1}(\simplify{{x2}-{x1}})$

$\phantom{A} = \dfrac{1}{2} (\var{y2}-\var{y1})(\simplify[zeroTerm]{{x2}-{x1}}) + \var{y1} (\simplify{{x2}-{x1}})$

The resulting area can be recognized as the sum of the area of a triangle with $\text{height}= (\simplify{{y2}-{y1}})$ and $\text{base}=(\simplify[zeroTerm]{{x2}-{x1}})$ and the area of a rectangle with $\text{height}=\var{y1}$ and the same base.

Alternatively, this further simplifies to the average height of the trapezoid times the base.

$A =\left( \dfrac{\var{y2}+\var{y1}}{2}\right) (\simplify[zeroTerm]{{x2}-{x1}})$

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This question is used in the following exam:

20. Integration Review by William Haynes in Engineering Statics.

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William Haynes 2 years, 5 months ago

William Haynes 2 years, 5 months ago

William Haynes 2 years, 5 months ago

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Generated value: `vector`

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Area of a trapezoid by integration

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

William Haynes 2 years, 5 months ago

William Haynes 2 years, 5 months ago

William Haynes 2 years, 5 months ago

Error in variable testing condition

Error

Warning

Generated value: vector

Parts

Gap-fill

Generated value: `vector`