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Thread: Intergration by substitution problem...

  1. #1
    Junior Member Bartimaeus's Avatar
    Jul 2006

    Intergration by substitution problem...

    Okay, I generally have no problems, except when the derivative of the term that needs to be substituted is a constant, say, 1 and the coefficient of the $\displaystyle dx$ or whatever in the original question is a pronumeral, say, $\displaystyle 3x$.

    Here is an example:

    Find the integral between $\displaystyle x=-3$ and $\displaystyle x=-2$ where $\displaystyle y=x(3+x)^7$ using the substitution $\displaystyle u=3+x$

    I have found that, by differentiation, $\displaystyle du=dx=1$

    My problem is:
    Once i get to this stage:

    $\displaystyle \int_{-3}^{-2}$ $\displaystyle (3+x)^7$ $\displaystyle x dx$

    $\displaystyle x=-2$ then $\displaystyle u=1$ and
    $\displaystyle x=-3$ then $\displaystyle u=0$

    and the $\displaystyle (3+x)^7$ then becomes $\displaystyle u^7$
    but what about the $\displaystyle x dx$?
    how do I get that into $\displaystyle du$?
    if the differentiated $\displaystyle u$ had an $\displaystyle x$ in it, then fine, I could do that, but since it doesn't, I'm in a pickle.

    And you can only have constant terms in front of the whole integral such as $\displaystyle 1/2 $$\displaystyle \int_{-3}^{-2}......$

    But what do I do for these types of integrals?
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  2. #2
    Kai is offline
    Junior Member
    Apr 2008
    Hi Bartimaeus,

    U are using the subsitution u=3+x, so x=u-3, and du=dx

    So finally u get to integrade u^7 *(u-3)du = u^8 - 3u^7 du ...
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  3. #3
    Super Member
    Jun 2008
    First of all: du = dx = 1 is nonsense.

    $\displaystyle \int x(3+x)^7~dx $

    Substituting u = 3 + x, we see that:
    x = u - 3
    dx = du

    There are no additional differentiated terms you need to multiply the original integrand with in order for you to change the variable of integration, so you can just change your terms and integrate.

    $\displaystyle \int x(3+x)^7~du = \int (u-3)u^7~du = \int u^8 - 3u^7~du$
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  4. #4
    Moo is offline
    A Cute Angle Moo's Avatar
    Mar 2008
    P(I'm here)=1/3, P(I'm there)=t+1/3

    Here is a way to do.

    $\displaystyle u=x+3$

    --> $\displaystyle \frac{du}{dx}=1 \implies {\color{blue}dx=du}$

    So when you're changing the integral, change dx with its formula with respect to du.

    Then, u=x+3 implies that x=u-3. So every x remaining in the integral will be substituted by this expression.

    $\displaystyle \int_{-3}^{-2} x(x+3)^7 ~dx=\int_{-3}^{-2} ({\color{red}u-3}) \cdot u^7 ~{\color{blue}du}=\int_{-3}^{-2} u^8-3u^7 ~du$

    Don't forget that the boundaries -3 and -2 apply to x and not to u.
    If you want to keep them, you will have to substitute back, that is to say when you got the integral in terms of u, substitute all the u's by x+3.
    Otherwise, you can change the boundaries so that they apply to u :
    x=-3 ---> u=0
    x=-2 ---> u=1
    The new boundaries are 0 and 1.
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