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Math Help - Definate Integral help please

  1. #1
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    Definate Integral help please

    I have maple and have used other tools online to find the answer, but I don't know the steps taken to get to it, I have a feeling its some algebra tricks that I'm just not remembering, but onto the problem.

    Its the integral of X/(sqrt(5+X)) from a=-1 to b=4, The indefinate integral is

    2 (-10 + x) Sqrt[5 + x]
    -----------------------
    3

    I just don't know how it got to that point, Please help!
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  2. #2
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    Quote Originally Posted by cloverz7 View Post
    I have maple and have used other tools online to find the answer, but I don't know the steps taken to get to it, I have a feeling its some algebra tricks that I'm just not remembering, but onto the problem.

    Its the integral of X/(sqrt(5+X)) from a=-1 to b=4, The indefinate integral is

    2 (-10 + x) Sqrt[5 + x]
    -----------------------
    3

    I just don't know how it got to that point, Please help!
    FOr \int_{-1}^{4} \frac{x}{\sqrt{x+5}}\,dx try the substitution u = x+5
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  3. #3
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    Or make the substitution t=\sqrt{x+5}

    ( dt=\frac{dx}{2t})
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  4. #4
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    If you do that how do you convert the top x to u? would you just add a -5 outside of the integral?
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  5. #5
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    sorry for the bump, but I really need some help
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  6. #6
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    Definate integral help please

    I have maple and have used other tools online to find the answer, but I don't know the steps taken to get to it, I have a feeling its some algebra tricks that I'm just not remembering, but onto the problem.

    Its the integral of X/(sqrt(5+X)) from a=-1 to b=4, The indefinate integral is

    2 (-10 + x) Sqrt[5 + x]
    -----------------------
    3

    I just don't know how it got to that point, Please help!


    I've tryed substituting u=5+x and even u=sqrt(5+x) I just can't seem to get anywhere with it, I think I'm doing soemthing wrong with converting the top x to u, Please Help meeeeeeee
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  7. #7
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    Hello, cloverz7!

    \int^4_{\text{-}1} \frac{x}{\sqrt{x+5}}\,dx \;=\;\frac{2}{3}(x-10)\sqrt{x+5}\,\bigg]^4_{\text{-}1}

    I don't know how it got to that point.
    Let: u \,=\,\sqrt{x+5}\quad\Rightarrow\quad x \,=\,u^2-5\quad\Rightarrow\quad dx \,=\,2u\,du

    Substitute: . \int\frac{u^2-5}{u}(2u\,du) \;=\; 2\int(u^2-5)\,du \;=\;2\left(\tfrac{1}{3}u^3 - 5u\right) \;=\;\tfrac{2}{3}u\left(u^2-15\right)

    Back-substitute: . \tfrac{2}{3}\sqrt{x+5}\,\bigg[\left(\sqrt{x+5}\right)^2 - 15\bigg] \;=\;\tfrac{2}{3}\sqrt{x+5}\,\bigg[x+5 - 15\bigg]

    And we have: . \frac{2}{3}(x-10)\sqrt{x+5}\,\bigg]^4_{\text{-}1} \qquad\hdots Got it?

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  8. #8
    Senior Member DeMath's Avatar
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    Quote Originally Posted by cloverz7 View Post
    I have maple and have used other tools online to find the answer, but I don't know the steps taken to get to it, I have a feeling its some algebra tricks that I'm just not remembering, but onto the problem.

    Its the integral of X/(sqrt(5+X)) from a=-1 to b=4, The indefinate integral is

    2 (-10 + x) Sqrt[5 + x]
    -----------------------
    3

    I just don't know how it got to that point, Please help!


    I've tryed substituting u=5+x and even u=sqrt(5+x) I just can't seem to get anywhere with it, I think I'm doing soemthing wrong with converting the top x to u, Please Help meeeeeeee
    Or

    \int\limits_{ - 1}^4 {\frac{x}{{\sqrt {x + 5} }}} dx = \int\limits_{ - 1}^4 {\frac{{x + 5 - 5}}{{\sqrt {x + 5} }}} dx = \int\limits_{ - 1}^4 {\left[ {\frac{{x + 5}}{{\sqrt {x + 5} }} - \frac{5}{{\sqrt {x + 5} }}} \right]} dx =

    = \int\limits_{ - 1}^4 {\left[ {\sqrt {x + 5}  - \frac{5}{{\sqrt {x + 5} }}} \right]} dx = \int\limits_{ - 1}^4 {\left[ {{{\left( {x + 5} \right)}^{{1 \mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2}}} - 5{{\left( {x + 5} \right)}^{{{ - 1} \mathord{\left/{\vphantom {{ - 1} 2}} \right.\kern-\nulldelimiterspace} 2}}}} \right]} dx =

    = \left. {\left[ {\frac{2}{3}{{\left( {x + 5} \right)}^{{3 \mathord{\left/{\vphantom {3 2}} \right.\kern-\nulldelimiterspace} 2}}} - 10{{\left( {x + 5} \right)}^{{1 \mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2}}}} \right]} \right|_{ - 1}^4 =  \ldots  =  - 12 - \left( { - \frac{{44}}{3}} \right) = \frac{8}{3}.
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cloverz7 View Post
    sorry for the bump, but I really need some help
    bumping is against the rules, and you have gotten help

    if you follow post #2's suggestion (yes, you must change the x to u - 5) then you end up with

    \int_4^9 \frac {u - 5}{\sqrt{u}}~du


    if you follow post #3's suggestion, then you end up with

    \int_2^3 2(t^2 - 5)~dt


    either is fine, it is up to you to do which you prefer. is there anything you do not understand on how we got these new integrals?
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