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Thread: Definite Integral Problem

  1. #1
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    Definite Integral Problem

    Attached is a problem I hand wrote. I don't get why on the first step, it has to be multiplied by -(1/2). is that part of a rule? Thanks!
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  2. #2
    Super Member Random Variable's Avatar
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    simple substitution

    $\displaystyle \int (5-2t)^{4} \ dt $

    let $\displaystyle u = 5-2t$

    then $\displaystyle du = -2 \ dt$

    which means that $\displaystyle dt = -\frac{du}{2} $

    so $\displaystyle \int (5-2t)^{4} \ dt = \int u^{4}\Big(-\frac{du}{2}\Big) = -\frac{u^{5}}{10} + C = -\frac{(5-2t)^{5}}{10} + C$
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  3. #3
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    I flipped through my textbook and it didn't say that it had to multiply (-1/2). How come it needs to multiply that? I'm still confused
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    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by fabxx View Post
    I flipped through my textbook and it didn't say that it had to multiply (-1/2). How come it needs to multiply that? I'm still confused
    You do know what the chain rule is? What we're looking for when we take

    $\displaystyle \int f(x)dx $ is some function $\displaystyle g(x) $ that when we differentiate it, it becomes $\displaystyle f(x) $. In other words, $\displaystyle g`(x) = f(x) $

    So to see why you need the -1/2 differentiate the answer and see what you get. You will notice that without the -1/2 you DO NOT obtain what was inside the integral!
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  5. #5
    Super Member Random Variable's Avatar
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    Because when you make the substitution, you switch from integrating with respect to $\displaystyle t$ to integrating with respect to $\displaystyle u$. To do that you need to know how to write $\displaystyle dt$ in terms of $\displaystyle du$. Have you covered integration by substitution in class?
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  6. #6
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    I understand that if I don't get (-1/2) it won't be the original problem and i do understand the chain rule.

    if its the chain rule

    should the anti differentiation of (-2t) = -2 [(t^2)/2] ?

    Thanks!
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  7. #7
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by fabxx View Post
    I understand that if I don't get (-1/2) it won't be the original problem and i do understand the chain rule.

    if its the chain rule

    should the anti differentiation of (-2t) = -2 [(t^2)/2] ?

    Thanks!
    $\displaystyle (5-2t)^4 $ this is what we want to get back to...

    If we differentiate

    $\displaystyle \frac { (5-2t)^5}{5} $ we get $\displaystyle -2(5-2t)^4 $ which is not what we originally had. Note that if we divide by -2 we end up with the same result, thus the anti derivative is

    $\displaystyle \frac { (5-2t)^5}{5} * - \frac{1}{2} $

    ...
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  8. #8
    MHF Contributor chiph588@'s Avatar
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    $\displaystyle (5-2t)^4 = 2^4(t-\tfrac52)^4 $

    So $\displaystyle \int (5-2t)^4dt = 2^4\int (t-\tfrac52)^4dt = \frac{2^4}{5}(t-\tfrac52)^5+C = \frac{2^5}{2\cdot5}(t-\tfrac52)^5+C =$ $\displaystyle \left(-\frac12\right)\cdot \frac{(5-2t)^5}{5}+C $
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