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Problem with Solution - Multivariable Calculus | MATH 2210, Assignments of Mathematics

Utah State University (USU)Mathematics

Prof. Bryan J. Bornholdt

Material Type: Assignment; Professor: Bornholdt; Class: MULTIVARI CALCULUS (QI)(H); Subject: Mathematics; University: Utah State University; Term: Unknown 1989;

Typology: Assignments

Pre 2010

Uploaded on 07/30/2009

koofers-user-2wk 🇺🇸

10 documents

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Math 2210 12.2 Problem 29

We may evaluate



1

0

1

0

35

dxdyeyx

xy

by integrating by parts (many, many times):

When integrating with respect to y, we choose

3

yu 

so that

dyydu

2

3

and

reversing the chain rule for dv with respect to y gives

)(

15 xy

exxv





. We see that

integration by parts in this case reduces the powers of both x and y. Thus, we have

 

 

   

 

   

 

 

 

 

 

 

 

5721

0)600000(3)60602771(

36060277

2433660277

182431836277

1218123618157

612943663

6663

663

63

3

01

1

0

2234

1

0

2234

1

0

2234

1

0

22234

1

0

232234

1

0

234

1

0

234

1

0

1

0

22334

1

0

1

0

222334

1

0

1

0

32334

1

0

1

0

2434

1

0

1

0

35

























































 

 

e

ee

xexxxx

dxexexxxx

dxexxexxxx

dxexxxexxxx

dxexxxxexxxx

dxxexxxx

dxxxeexexex

dxxeyexeyxeyx

dxdyexyexeyxeyx

dxdyyexeyxeyx

dxdyeyxeyxdxdyeyx

x

xx

x

xxxx

xyxyxyxy

xyxyxy

Thank you for reading. (Who needs a computer for this?)

Discover Assignments of Mathematics Utah State University (USU)

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Download Problem with Solution - Multivariable Calculus | MATH 2210 and more Assignments Mathematics in PDF only on Docsity!

Math 2210 12.2 Problem 29

We may evaluate 

1 0 1 0

x^5 y^3 exy^ dydx

by integrating by parts ( many, many times ):

When integrating with respect to y, we choose u^ ^^ y^3 so that du^ ^3 y^2 dy and

reversing the chain rule for dv with respect to y gives (^ )

v  x^5 x ^1 exy

. We see that

integration by parts in this case reduces the powers of both x and y. Thus, we have

1 0 1 0 4 3 2 2 1 0 4 3 2 2 1 0 4 3 2 2 1 0 4 3 2 2 2 1 0 4 3 2 2 3 2 1 0 4 3 2 1 0 4 3 2 1 0 1 0 4 3 3 2 2 1 0 1 0 4 3 3 2 2 2 1 0 1 0 4 3 3 2 3 1 0 1 0 4 3 4 2 1 0 1 0 5 3  

e e e x x x x e x x x x x e x e dx x x x x e x x e dx x x x x e x x x e dx x x x xe x x x x e dx x x x xe x dx xe xe xe xe x dx xye xye xye xe dx xye xye xye xe dy dx xye xye xye dy dx xye dydx xye xye dy dx x x x x x x x x x x x x x x xy xy xy xy xy xy xy xy xy xy xy xy xy xy

Problem with Solution - Multivariable Calculus | MATH 2210, Assignments of Mathematics

Related documents

Partial preview of the text

Download Problem with Solution - Multivariable Calculus | MATH 2210 and more Assignments Mathematics in PDF only on Docsity!

Math 2210 12.2 Problem 29

We may evaluate 

x^5 y^3 exy^ dydx

by integrating by parts ( many, many times ):

When integrating with respect to y, we choose u^ ^^ y^3 so that du^ ^3 y^2 dy and

reversing the chain rule for dv with respect to y gives (^ )

v  x^5 x ^1 exy

. We see that

integration by parts in this case reduces the powers of both x and y. Thus, we have

Thank you for reading. (Who needs a computer for this?)