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Math Help - Integrating a function that's proportional to...

  1. #1
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    Integrating a function that's proportional to...

    I am supposed to integrate a function that is proportional to
    (10+x)^{-2}
    My initial strategy was to treat that as the reciprocal of
    x^2 + 20x + 100
    but that leaves me with the whole thing raised to an exponent of -1. I don't know how to take an antiderivative anything raised to -1!

    So I decided to make a new variable, A, and set that equal to x+10. Taking the antiderivative of
    A^{-2}
    was pretty easy, giving me
    -1/A = -(x+10)^{-1}
    I just don't know if what I did was mathematically correct. (Getting a definite integral at that point was easy, and resulted in an answer close to one of my multiple choices.)

    The thing is, I'm treating the whole thing as if I were getting the antid. of
    (10+x)^{-2}.
    Really I'm just dealing with an unknown function proportional to that. Is this something I haven't learned yet

    or can I just take the antiderivative of what my function is proportional to?
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  2. #2
    Moo
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    Hello,

    I didn't get the whole problem oO

    But the antiderivate of (10+x)^{-2} is easy to find.

    An antiderivative of u'(x)u^n(x) is \frac{u^{n+1}(x)}{n+1}

    Here, u(x)=10+x
    u'(x)=1

    So (10+x)^{-2}=1*(10+x)^{-2}=u'(x)u^{-2}(x)

    As you've calculated, an antiderivative will be -u^{-1}(x)=\frac{-1}{10+x}

    What do you mean by proportional ? You have a function equal to :

    f(x)=a (10+x)^{-2} with a a constant ?
    If so, it's ok, the antiderivative doesn't take care of the constants...

    An antiderivative of f will be a*[antiderivative of 1/(10+x)^2]
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  3. #3
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    Okay, thanks for helping me with the antiderivative, Moo. Here is the rascally problem in its entirety:

    "The lifetime of a part has a continuous distribution on the interval (0, 40) with probability density function f, where f(x) is proportional to
    (10 + x)^{-2}

    Calculate the probability that the lifetime of the part is less than 6."

    I've never seen the "proportional to..." language in any other problem, so I'm wondering if I missed something. I am betting that all all density function proportional to the same thing can be translated into a single distribution function. Unfortunately, this requires a deeper understanding of calculus than I have yet developed.
    Last edited by Boris B; April 13th 2008 at 12:34 PM.
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    Okay, I figured it out. You pointed out my antiderivative was correct. In order to get from that to the actual distribution function, we had to correct by a factor: the proportionality constant. I had forgotten to take into account the fact that the integral (0 to 40) f(x) dx = 1, because it's a probability question.
    C \int_0^{40} f(x) dx = 1
    1 = \int_0^{40} C (10+x)^{-2} dx = -C (10+x)^{-1} = C/10 - C/50
    My answer was off by exactly that factor.
    Last edited by Boris B; April 14th 2008 at 12:14 AM.
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  5. #5
    Moo
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    I'm sorry I can't check it, it's out of my abilities
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  6. #6
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    Quote Originally Posted by Boris B View Post
    Okay, I figured it out. You pointed out my antiderivative was correct. In order to get from that to the actual distribution function, we had to correct by a factor: the proportionality constant. I had forgotten to take into account the fact that the integral (0 to 40) f(x) dx = 1, because it's a probability question.
    C \int_0^40 f(x) dx = 1
    1 = \int_0^40 C (10+x)^{-2} dx = -C (10+x)^{-1} = C/10 - C/50
    My answer was off by exactly that factor.
    I don't quite follow this. The term "0C" is going to make the term you want to integrate 0, thus the integral will be 0.  \int _{a}^{b} 0dx=0 for all a,b in R I'm pretty sure.
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  7. #7
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    Oh dear, my bad, I didn't enclose my 40s in braces, so it put 4s in superscript and left the 0 floating around. I fixed it above.

    I've pretty much figured this creature out now.
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  8. #8
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    Yeah that looks good now. Proportional to a function just means some constant times that function like you said. Good work.
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