To evaluate the double integral $$ int limits_0^8 { left( { int limits_{y/2}^{ left( {y/2} ight) + 1} { left( {{{2x - y} over 2}} ight)dx} } ight)dy, , ,} $$ we make the substitution $$u = left( {{{2x - y} over 2}} ight)$$ and $$v = {y over 2}.$$ The integral will reduce to

$$ int limits_0^4 { left( { int limits_0^1 {2udu} } ight)dv} $$

$$ int limits_0^4 { left( { int limits_0^2 {2udu} } ight)dv} $$

$$ int limits_0^4 { left( { int limits_0^1 {udu} } ight)dv} $$

$$ int limits_0^4 { left( { int limits_0^{21} {2udu} } ight)dv} $$

The double integral ,$$ , int_0^a { int_0^y {f left( {x,y} ight) ,dx ,dy , , ,} }$$ is equivalent to

$$ int_0^a { int_x^a {f left( {x,y} ight)dx ,dy} } $$

$$ int_0^x { int_0^y {f left( {x,y} ight)dx ,dy} } $$

$$ int_0^a { int_x^y {f left( {x,y} ight)dx ,dy} }$$

$$ int_0^a { int_0^a {f left( {x,y} ight)dx ,dy} } $$

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Double Integral

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Question 1

what is not the property of double integrals?

linearity
area property
commutativity
additivity

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Question 2

To evaluate the double integral [Tex]\int\limits_0^8 {\left( {\int\limits_{y/2}^{\left( {y/2} \right) + 1} {\left( {{{2x - y} \over 2}} \right)dx} } \right)dy,\,\,} [/Tex] we make the substitution [Tex]u = \left( {{{2x - y} \over 2}} \right)[/Tex] and [Tex]v = {y \over 2}.[/Tex] The integral will reduce to

[Tex]\int\limits_0^4 {\left( {\int\limits_0^2 {2udu} } \right)dv} [/Tex]
[Tex]\int\limits_0^4 {\left( {\int\limits_0^1 {2udu} } \right)dv} [/Tex]
[Tex]\int\limits_0^4 {\left( {\int\limits_0^1 {udu} } \right)dv} [/Tex]
[Tex]\int\limits_0^4 {\left( {\int\limits_0^{21} {2udu} } \right)dv} [/Tex]

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Question 3

The double integral \,[Tex]\,\int_0^a {\int_0^y {f\left( {x,y} \right)\,dx\,dy\,\,\,} }[/Tex] is equivalent to

[Tex]\int_0^x {\int_0^y {f\left( {x,y} \right)dx\,dy} } [/Tex]
[Tex]\int_0^a {\int_x^y {f\left( {x,y} \right)dx\,dy} }[/Tex]
[Tex]\int_0^a {\int_x^a {f\left( {x,y} \right)dx\,dy} } [/Tex]
[Tex]\int_0^a {\int_0^a {f\left( {x,y} \right)dx\,dy} } [/Tex]

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Question 4

If the value of the double integral

[Tex]\int_{x=3}^{4} \int_{y=1}^{2} \frac{dydx}{(x + y)^2}[/Tex]

is [Tex]\log_e(\frac{a}{24})[/Tex], then [Tex]a[/Tex] is __________ (answer in integer).

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Question 5

The double integral ∬_Rf(x,y) dA represents:

Area of a curve
Volume under the surface z = f(x,y)
Length of a curve
Slope of a plane

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