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Math Help - Iterated Integrals

  1. #1
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    Iterated Integrals

    1. The problem statement, all variables and given/known data

    Evaluate the iterated integral

    2. Relevant equations

    Fundamental Theorem of Calculus & Fubini's Theorem

    3. The attempt at a solution

    I have been working on this problem for the last hour and haven't been able to solve it thus far. I integrated the inside of the integral with respect to y from 4 to 3, and then integrated the result from the first integral with respect to x from 2 to 1. In the process, I used substitution to solve the integrals.

    The answer I keep on getting is (1/(4(3x+4)^4)) - (1/(4(3x+3)^4)) and solve from x= 2 to 1 which results in a small fractional answer of 7.5674005*10^(-5). But my online program keeps on saying its incorrect.

    Can anyone give me a heads up as to where I am going wrong? This is one of the few iterated integrals that is giving me problems right now
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  2. #2
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    Nov 2009
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    suggestion

    Quote Originally Posted by skaterbasist View Post
    1. The problem statement, all variables and given/known data

    Evaluate the iterated integral

    2. Relevant equations

    Fundamental Theorem of Calculus & Fubini's Theorem

    3. The attempt at a solution

    I have been working on this problem for the last hour and haven't been able to solve it thus far. I integrated the inside of the integral with respect to y from 4 to 3, and then integrated the result from the first integral with respect to x from 2 to 1. In the process, I used substitution to solve the integrals.

    The answer I keep on getting is (1/(4(3x+4)^4)) - (1/(4(3x+3)^4)) and solve from x= 2 to 1 which results in a small fractional answer of 7.5674005*10^(-5). But my online program keeps on saying its incorrect.

    Can anyone give me a heads up as to where I am going wrong? This is one of the few iterated integrals that is giving me problems right now
    what is the correct answer???i got 1/12 although i am not sure whether that is correct.
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  3. #3
    MHF Contributor

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    \int_1^2\int_3^4 (3x+ y)^{-2}dydx
    Let u= 3x+y so that dy= du. When y= 3, u= 3x+3 and when y= 4, u= 3x+4. The integral becomes:
    \int_{x=1}^2\int_{u= 3x+3}^{3x+4} u^{-2}du= -\int_1^2\left(\frac{1}{3x+4}- \frac{1}{3x+ 3}\right)dx
    = -\int_1^2 \frac{1}{3x+4}dx+ \int_1^2\frac{1}{3x+3}dx

    In the first integral, let v= 3x+ 4 so that dx=(1/3)dv. When x= 1, v= 7 and when x= 2, v= 10. That integral becomes \frac{1}{3}\int_7^10 \frac{1}{v} dv= \frac{1}{3}(ln(10)- ln(7))= \frac{1}{3}ln(10/7)

    In the second integral, let w= 3x+ 3 so that dx= (1/3)dw. When x= 1, w= 6 and when x= 2, w= 9. That integral becomes \frac{1}{3}\int_6^9 \frac{1}{w}dw= \frac{1}{3}(ln(9)- ln(6))= \frac{1}{3}ln(3/2).

    The original integral is -\frac{1}{3}\left(ln(10/7)- ln(3/2)\right)= -\frac{1}{3}ln(20/21)
    That is about 0.01626.
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  4. #4
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    Thank you. I made the stupid mistake of differentiating u^-2 when I should have been integrating (for the substitution of the integral with respect to y)
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