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Thread: 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 $\displaystyle \int_{-1}^0(2x+3)^2dx$, but I don't get the right answer unless I square the $\displaystyle (2x+3)$ and then proceed normally.

    What about substitution on $\displaystyle \int_3^2\frac{x^2-1}{x-1}dx$?
    Last edited by cinder; Apr 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 $\displaystyle \int_{-1}^0(2x+3)^2dx$, but I don't get the right answer unless I square the $\displaystyle (2x+3)$ and then proceed normally.
    What about substitution on $\displaystyle \int_3^2\frac{x^2-1}{x-1}dx$?
    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:

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

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

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

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

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

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

    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:

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

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

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

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

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

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

    From here on you know what to do!

    Greetings

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

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

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

    then

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

    and solve normally, just using $\displaystyle du$?
    Last edited by cinder; Apr 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 $\displaystyle du$?
    Yes, that what $\displaystyle du$ means, a function in terms of $\displaystyle u$ whose derivative gives you whatever the function is. Or another way of saying the antiderivative in terms of $\displaystyle u$. 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 $\displaystyle du$ means, a function in terms of $\displaystyle u$ whose derivative gives you whatever the function is. Or another way of saying the antiderivative in terms of $\displaystyle u$. 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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