Double Integrals Calculator - Symbolab 6q411t

4/3/2015

Double Integrals Calculator ­ Symbolab Solutions

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Double Integrals Calculator Solve double integrals step by step Enter a topic

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∫0 ∫0

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xe dxdy

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Double Integrals Triple Integrals Multiple Integrals Derivatives Derivative Applications

ln ( 3 ) 1 xy ∫0 xe dxdy =   2 ln ( 3 )

∫0

« Hide Steps

−1

Steps

In the previous post we covered th

Series ODE (new) Laplace Transform (new)

Advanced Math Solution Integral Calculator, commo functions

integration rules (click here).  Befo

ln ( 3 ) 1 xy ∫0 xe dxdy

∫0

continue with more advanced tech will cover some common integrals e...

Statistics Hide Steps

y 1 ∫0 xexydx =  e ( y −21 ) + 1

y

1

∫0 xexydx xy

xy

y

y

Hide Steps

∫ xexydx =  xe −  e 2 + C

Compute the indefinite integral:

Advanced Math Solution Integral Calculator, the bas

Integration is the inverse of differe

∫ xexydx

Even though derivatives are fairly s

forward, integrals are not. Some in Apply Integration By Parts:

problems require techniques such

∫ uv' = uv − ∫ u'v yx

yx

u = x, u' = 1, v' = e , v =  e

y

yx

yx

= x  e − ∫ 1  e dx y y xy

=  xe

y

xy

− ∫  e dx y

xy

xy

y

y

Hide Steps

∫  e dx =  e 2 xy

∫  e dx y

Take the constant out:

∫a · f ( x ) dx = a · ∫f ( x ) dx

xy

=  1 ∫ e dx y

u = xy:

du = ydx,

∫f ( g ( x ) ) · g' ( x ) dx = ∫f ( u ) du,

Apply Integral Substitution:

u = g(x)

dx =  1 du y

u =  1 ∫ e  1 du y y u

=  1 ∫  e du y

y

Take the constant out:

∫a · f ( x ) dx = a · ∫f ( x ) dx

u =  1  1 ∫ e du y y

http://www.symbolab.com/solver/double­integrals­calculator/%5Cint_%7B0%7D%5E%7Bln3%7D%5Cint_%7B0%7D%5E%7B1%7D%20xe%5E%7Bxy…

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4/3/2015

Double Integrals Calculator ­ Symbolab Use the common integral:

=  1  1 e

∫ eudu = eu

u

y y

Substitute back

=  1  1 e

u = xy

xy

y y

Simplify xy

=  e

2

y

xy

xy

=  xe

−  e

y

2

y

Add a constant to the solution xy

xy

=  xe

−  e

y

2

y

+C

b ∫a f ( x ) dx = F ( b )

(

  y

(

  y

xy

y

( ) y

− F ( a ) = limx→ b− ( F ( x ) ) − limx→ a+ ( F ( x ) )

xy

limx→ 0+ xe

Hide Steps

y 1 ∫0 xexydx =  e ( y 2− 1 ) − −  12

Compute the boundaries:

−e

  2 y

)

= −  1 2

)

y =  e ( y − 1 )

y

= −  1 2 y xy

xy

limx→ 1− xe

−e

  2 y

2

y

y

=  e ( y − 1 ) 2

y

y =  e ( y − 1 ) − − 1

( )  2 y

2

y

y =  e ( y − 1 ) − − 1

( )  2 y

2

y

Simplify y =  e ( y − 1 ) + 1 2

y

ln ( 3 )

= ∫0

ln ( 3 )

∫0

(

(

)

y

e ( y − 1 ) + 1 dy   2 y Hide Steps

)

y

e ( y − 1 ) + 1 dy = 2 − 1  ln ( 3 )   2 y

ln ( 3 ) ey ( y − 1 ) + 1 dy   2

∫0

y

Compute the indefinite integral:

y y ∫  e ( y −21 ) + 1 dy = ey −  e ( y − 1 ) −  1 + C

y

y

Hide Steps

y

y

∫  e ( y −21 ) + 1 dy y

Apply the Sum Rule:

∫f ( x ) ± g ( x ) dx = ∫f ( x ) dx ± ∫g ( x ) dx

y = ∫  e ( y − 1 ) dy + ∫  1 dy 2

y

2

y

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4/3/2015

Double Integrals Calculator ­ Symbolab Hide Steps

y y ∫  e ( y 2− 1 ) dy = ey −  e ( y − 1 )

y

y

y

∫  e ( y 2− 1 ) dy y

∫ uv' = uv − ∫ u'v

Apply Integration By Parts:

y

y

u = e ( y − 1 ) , u' = e y, v' =  1 , v = −  1 y

2

y

(  y )

y

y

(  y )

= e ( y − 1 ) − 1 − ∫ e y − 1 dy y y = −  e ( y − 1 ) − ∫ − e dy y

Hide Steps

∫ − eydy = −ey ∫ − eydy ∫a · f ( x ) dx = a · ∫f ( x ) dx

Take the constant out:

y

= −∫ e dy

Use the common integral:

= −e

∫ eydy = ey

y

y y = −  e ( y − 1 ) − −e

(

y

)

Simplify y y = e −  e ( y − 1 ) y Hide Steps

∫  12 dy = −  1 y

y

∫  12 dy y

−2 1  2 =y y

=∫ y

−2

dy

Apply the Power Rule:

=  y

∫xadx =  x

a+1

a+1

,

a ≠ −1

−2+1

−2 + 1

Simplify

= −  1 y y y = e −  e ( y − 1 ) −  1 y y

Add a constant to the solution y y = e −  e ( y − 1 ) −  1 + C y y

Compute the boundaries:

ln ( 3 )

∫0

(

y

)

e ( y − 1 ) + 1 dy = 2 − 1  ln ( 3 )   2 y

Hide Steps

b

∫a f ( x ) dx = F ( b ) − F ( a ) = limx→ b− ( F ( x ) ) − limx→ a+ ( F ( x ) )

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4/3/2015

Double Integrals Calculator ­ Symbolab

(

)

y y limy→ 0+ e − e ( y − 1 ) − 1 = 1

 

y

 y

=1

(

)

y y limy→ln ( 3 ) − e − e ( y − 1 ) − 1 =   2    y y ln ( 3 )

=  2

ln ( 3 )

=  2

ln ( 3 )

=  2

ln ( 3 )

−1

Notes add notes here

−1

=   2 −1 ln ( 3 )

Save

Examples

∫∫

− 1 dxdx  x

2 1 1 ∫0 ∫0   x 2 dydx

1+y

∫∫x

2

1 1

∫0 ∫0 xy dydx  

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