Evaluate the iterated integral (Calculus 3 double integrals)
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Evaluate the iterated integral (Calculus 3 double integrals)

[From: ] [author: ] [Date: 12-11-19] [Hit: ]
= (1/3) [ln 10 - ln 11 - ln 7 + ln 8].I hope this helps!......
Evaluate the iterated integral

http://imageshack.us/photo/my-images/801/1c99b14df9f17164197fea0.png/

im unsure of how to integrate this any help would be great!

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Use Fubini's Theorem.

∫(x = 2 to 3) ∫(y = 1 to 2) (3x + y)^(-2) dy dx
= ∫(x = 2 to 3) -(3x + y)^(-1) {for y = 1 to 2} dx, integrating with respect to y
= ∫(x = 2 to 3) [1/(3x + 1) - 1/(3x + 2)] dx
= [(1/3) ln |3x + 1| - (1/3) ln |3x + 2|] {for x = 2 to 3}
= (1/3) [ln |3x + 1| - ln |3x + 2|] {for x = 2 to 3}
= (1/3) [ln 10 - ln 11 - ln 7 + ln 8].

I hope this helps!
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keywords: integrals,iterated,double,the,integral,Calculus,Evaluate,Evaluate the iterated integral (Calculus 3 double integrals)
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