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U sub of a real double integral

  1. Jul 26, 2012 #1
    1. The problem statement, all variables and given/known data


    ∫∫ 1 / (2x + 3y), R = [0,1] x [1,2]



    2. Relevant equations

    - Iterated integrals
    - u sub

    3. The attempt at a solution

    Here is my attempt at solving this (I must be screwing up on the algebra)

    Integrating with respect to x
    u = 2x, dy = 2

    u^-2 du

    u^-2 = - (1/u) [double chcked with wolfram alpha]
    ∫∫ (2x + 3y) ^ -2 -> -2 ∫ -(1/u) (2x + 3y) ^ -1

    This is the part that stumps me....Sorry if I'm being vague. This is from Paul Online Notes Calculus III (http://tutorial.math.lamar.edu/Classes/CalcIII/IteratedIntegrals.aspx#MultInt_Iterated_Ex1d)

    THe part I want to understand is how they got -1/2 * ...... (2x+3y) ^ -1

    Thanks!
     
  2. jcsd
  3. Jul 26, 2012 #2

    LCKurtz

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    Gold Member

    For starters, you didn't copy the problem correctly. Should be$$
    \int_1^2\int_0^1 \frac 1 {(2x+3y)^2}\, dxdy$$
    Let ##u=2x+3y## and ##du=2dx##. y is constant for the inner integral. The inner integral becomes$$
    \int u^{-2} \frac 1 2 du = \frac 1 2 \frac {u^{-1}}{-1}$$
     
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