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Math Help - 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

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

    let  u = 5-2t

    then  du = -2 \ dt

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

    so  \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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  4. #4
    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

    \int f(x)dx is some function  g(x) that when we differentiate it, it becomes  f(x) . In other words,  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 t to integrating with respect to u. To do that you need to know how to write  dt in terms of 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!
     (5-2t)^4 this is what we want to get back to...

    If we differentiate

    \frac { (5-2t)^5}{5} we get  -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

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

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

    So  \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 =  \left(-\frac12\right)\cdot \frac{(5-2t)^5}{5}+C
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