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Math Help - Definite Intergrals problem

  1. #1
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    Definite Intergrals problem

    I'm only posting here because I am not seeing any way to do this in my Text book and my two math help books.

    Let \(\displaystyle \int_{10}^{19} f(x) dx =7, \ \int_{10}^{13} f(x) dx=9, \ \int_{16}^{19} f(x)dx =10\).

    Find \(\displaystyle \int_{13}^{16} f(x)dx=\)

    and  \(\displaystyle \int_{16}^{13} (7 f(x)- 9)dx=\)
    I'm stuck on the first one.

    I'm trying to use the Additive Interval Property as Shown Below.

    \mathbf{\int_{a}^{b}f(x) dx= \int_{c}^{a}f(x) dx + \int_{c}^{b}f(x) dx}
    Where a= 13, b=16.


    So I try:
    \int_{10}^{13} f(x) dx + \int_{16}^{19} f(x)dx
    But keep ending up with the wrong answer, I've resorted by trying different combinations but with no success (even substitution).
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  2. #2
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    Try \displaystyle \int _{10}^{19} f(x)\,dx - \int_{16}^{19}f(x)\,dx - \int_{10}^{13}f(x)\,dx.
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  3. #3
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    Well that worked...but why?
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  4. #4
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    Think about the required interval that is [13,16]

    Danny's answer took a bigger interval [10,19] and cut off a bit at the bottom of this interval [10,13] and a bit at the top [16,19] leaving you with [13,16].

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  5. #5
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     \displaystyle \int_{16}^{13} (7 f(x)- 9)~dx

     \displaystyle \int_{16}^{13} 7 f(x)~dx-\int_{16}^{13} 9~dx

     \displaystyle 7\int_{16}^{13}  f(x)~dx-\int_{16}^{13} 9~dx

    You can use the result to the first question to help finish this.
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  6. #6
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    You have

    \displaystyle \int_{10}^{19} f(x)\,dx,\;\;\;\int_{10}^{13} f(x)\,dx,\;\;\;\int_{16}^{19} f(x)\,dx

    You want

    \displaystyle \int_{13}^{16} f(x)\,dx.

    \displaystyle \int_{10}^{19} f(x)\,dx = \int_{10}^{13} f(x)\,dx + \int_{13}^{16} f(x)\,dx + \int_{16}^{19} f(x)\,dx

    Now isolate what you want (you have the other three integral values).
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  7. #7
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    Quote Originally Posted by pickslides View Post
     \displaystyle 7\int_{16}^{13}  f(x)~dx-\int_{16}^{13} 9~dx

    You can use the result to the first question to help finish this.
    Okay well I have done that.

    At first I tried.

    7(-12)-9=-93

    That was wrong.

    So I then tried

    7(-12)-9(-12)=24

    And that was wrong.

    I'll keep looking at this rule:

    \int_{a}^{b}[f(x)\pm g(x)]dx= \int_{a}^{b}f(x)dx \pm \int_{a}^{b} g(x)dx
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  8. #8
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    Also \displaystyle \int_{a}^{b}f(x)~dx = -\int_{b}^{a}f(x)~dx
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  9. #9
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    Just a quick question what's dx supposed to be?

    Even with the rule you just said above.
    I get the same answers as I did before only the sign is changed.
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  10. #10
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    It's how to do the second part I can't figure out at the moment now.

    I was looking at the wrong formula too.

    I need to be looking at :

    \int_{a}^{b} kf(x)dx=k \int_{a}^{b} f(x)dx

    I'm starting to think that this has something to do with chain rule or compound functions (something along those lines).
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  11. #11
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    Quote Originally Posted by Zanderist View Post
    Just a quick question what's dx supposed to be?
    That the integration is with respect to x.
    I'm starting to think that this has something to do with chain rule or compound functions (something along those lines).
    It's to do with the linearity of integration: if k is any constant, then \displaystyle \int_{a}^{b} kf(x)dx=k \int_{a}^{b} f(x)dx.

    It's called the constant factor rule of integration.
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  12. #12
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    I've just solved it!

    \(\displaystyle \int_{16}^{13} (7 f(x)- 9)dx=\)
    <br />
-7 \int_{16}^{13} f(-12) + 9\int_{16}^{13}dx
    The dx becomes b-a, or 16-13 = 3

    so you end up with
    -7(-12)+9(3)=111
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