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

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

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

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

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


    So I try:
    $\displaystyle \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 $\displaystyle \displaystyle \int _{10}^{19} f(x)\,dx - \int_{16}^{19}f(x)\,dx - \int_{10}^{13}f(x)\,dx$.
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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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    $\displaystyle \displaystyle \int_{16}^{13} (7 f(x)- 9)~dx$

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

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

    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

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

    You want

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

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

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

    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:

    $\displaystyle \int_{a}^{b}[f(x)\pm g(x)]dx= \int_{a}^{b}f(x)dx \pm \int_{a}^{b} g(x)dx$
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  8. #8
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    Also $\displaystyle \displaystyle \int_{a}^{b}f(x)~dx = -\int_{b}^{a}f(x)~dx$
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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 :

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

    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 $\displaystyle x$.
    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 $\displaystyle k$ is any constant, then $\displaystyle \displaystyle \int_{a}^{b} kf(x)dx=k \int_{a}^{b} f(x)dx.$

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

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