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Math Help - substitution method question

  1. #1
    Junior Member cinder's Avatar
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    substitution method question

    Do can you tell when you can use the substitution method? I just tried it on \int_{-1}^0(2x+3)^2dx, but I don't get the right answer unless I square the (2x+3) and then proceed normally.

    What about substitution on \int_3^2\frac{x^2-1}{x-1}dx?
    Last edited by cinder; April 9th 2006 at 10:40 PM.
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  2. #2
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    Quote Originally Posted by cinder
    Do can you tell when you can use the substitution method? I just tried it on \int_{-1}^0(2x+3)^2dx, but I don't get the right answer unless I square the (2x+3) and then proceed normally.
    What about substitution on \int_3^2\frac{x^2-1}{x-1}dx?
    Hello,

    to 1.:
    use u(x) = 2x+3. Then du/dx = 2. That means du = 2*dx.

    So you could use the substitution method if there are the factor 2:

    \int_{-1}^0 \frac{1}{2} \cdot 2 \cdot (2x+3)^2dx=\int_{-1}^0 \frac{1}{2} u^2 du

    \int_{-1}^0 \frac{1}{2} u^2 du= \frac{1}{6}u^3+c

    Now you can re-substitute and you'll get:

    \int_{-1}^0 \frac{1}{2} \cdot 2 \cdot (2x+3)^2dx=\frac{1}{6} \cdot (2x+3)^3+c

    to 2.:
    Factorize (x^2-1) = (x+1)(x-1). Then you can cancel (x-1) and you'll get:

    \int_3^2\frac{x^2-1}{x-1}dx= \int_3^2 (x+1)dx.

    From here on you know what to do!

    Greetings

    EB
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  3. #3
    Junior Member cinder's Avatar
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    Quote Originally Posted by earboth
    Hello,

    to 1.:
    use u(x) = 2x+3. Then du/dx = 2. That means du = 2*dx.

    So you could use the substitution method if there are the factor 2:

    \int_{-1}^0 \frac{1}{2} \cdot 2 \cdot (2x+3)^2dx=\int_{-1}^0 \frac{1}{2} u^2 du

    \int_{-1}^0 \frac{1}{2} u^2 du= \frac{1}{6}u^3+c

    Now you can re-substitute and you'll get:

    \int_{-1}^0 \frac{1}{2} \cdot 2 \cdot (2x+3)^2dx=\frac{1}{6} \cdot (2x+3)^3+c

    to 2.:
    Factorize (x^2-1) = (x+1)(x-1). Then you can cancel (x-1) and you'll get:

    \int_3^2\frac{x^2-1}{x-1}dx= \int_3^2 (x+1)dx.

    From here on you know what to do!

    Greetings

    EB
    Okay, so on the first one, could you do:

    \int_{-1}^0(2x+3)^2dx

    u = (2x + 3)
    du = 2dx which becomes \frac{1}{2}du=dx

    then

    \int_1^3u^2\frac{1}{2}du = \frac{1}{2}\int_1^3u^2du

    and solve normally, just using du?
    Last edited by cinder; April 10th 2006 at 06:13 AM.
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  4. #4
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    Quote Originally Posted by cinder
    and solve normally, just using du?
    Yes, that what du means, a function in terms of u whose derivative gives you whatever the function is. Or another way of saying the antiderivative in terms of u. That is the entire point of substitution because it simplifys the integrand into a simpler one.
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  5. #5
    Junior Member cinder's Avatar
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    Quote Originally Posted by ThePerfectHacker
    Yes, that what du means, a function in terms of u whose derivative gives you whatever the function is. Or another way of saying the antiderivative in terms of u. That is the entire point of substitution because it simplifys the integrand into a simpler one.
    Okay, did I do the substitution right?
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  6. #6
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    Quote Originally Posted by cinder
    Okay, did I do the substitution right?
    Yes
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  7. #7
    Junior Member cinder's Avatar
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    Quote Originally Posted by ThePerfectHacker
    Yes
    Thanks.
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