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Math Help - exponent rule integrate

  1. #1
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    exponent rule integrate

    The value of is
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xcelxp View Post
    The value of is
    so i don't really see why you are having problems with this one, you simply use the power rule. i've seen you do far more complicated integrals than this.

    int{6:10}[1/x^2]dx = int{6:10}[x^-2]dx
    = [-x^-1] between [6 10]
    = -10^-1 + 6^-1
    = 1/15
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  3. #3
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    Integrating is simple, I just have no idea what to do with the 6 & 10
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xcelxp View Post
    Integrating is simple, I just have no idea what to do with the 6 & 10
    ok, let me explain

    if we have the int{f(x)}dx = F(x), that is F(x) is the integral of f(x), then:

    int{f(x)}dx from a going up to b = F(b) - F(a)

    this is called the fundamental theorem of calculus.

    so you do the integration as normal. when you end up with the final expression, you plug in the number at the top of the integral sign and subtract the expression when you plug in the number at the bottom of the integral sign.

    so in this problem we saw the integral was -x^-1 (when finding the definate integral, that is integral between two given limits, we can forget about the + C since they'll cancel out anyway)

    so if i let F(x) = -x^-1, then the integral between 6 and 10 is given by

    F(10) - F(6)

    and that's how i got -10^-1 - (-6^-1) = -10^-1 + 6^-1 = 1/15

    so look up in your text book for the "fundamental theorem of calculus" it tells you how to deal with these problems
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  5. #5
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    thats simple enough, thanks.

    Find the area enclosed between and from to .




    Now this one is driving me crazy.
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  6. #6
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by xcelxp View Post


    Find the area enclosed between and from to .
    For this, we integrate (f(x) - g(x))dx from x = -5 to x = 6

    INT {-5,6} [(.3x^2 + 3) - (x)] dx
    INT {-5,6} [.3x^2 - x + 3] dx
    [.1x^3 - .5x^2 + 3x] | {-5,6}
    (.1(-5)^3 - .5(-5)^2 + 3(-5)) - (.1(6)^3 - .5(6)^2 + 3(6))

    You should have no problem with the arithmatic.

    Edit!
    [.1x^3 - .5x^2 + 3x] | {-5,6} = (.1(6)^3 - .5(6)^2 + 3(6)) - (.1(-5)^3 - .5(-5)^2 + 3(-5))

    Thanks, Jhevon
    Last edited by ecMathGeek; April 18th 2007 at 02:37 PM.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    For this, we integrate (f(x) - g(x))dx from x = -5 to x = 6

    INT {-5,6} [(.3x^2 + 3) - (x)] dx
    INT {-5,6} [.3x^2 - x + 3] dx
    [.1x^3 - .5x^2 + 3x] | {-5,6}
    (.1(-5)^3 - .5(-5)^2 + 3(-5)) - (.1(6)^3 - .5(6)^2 + 3(6))

    You should have no problem with the arithmatic.
    ecMathGeek is right! (though he did plug in the limits incorrectly). to find the area between two curves it is the integral of (higher curve - lower curve) between the desired limits

    this one isn't so bad, look out for problems where the curves switch roles! that is the higher curve in one interval, is the lower curve in the other interval and you have to integrate across both intervals, in that case, you would split the integral in two with the respective limits
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