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Math Help - Rate of oil production..

  1. #1
    Newbie melanie's Avatar
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    Rate of oil production..

    Here's a problem. Not sure if I should just sub in or firstly integrate and then sub in. Please help..

    an oil company found that the rate of oil production from an oil well is given by a function of time t, in years:

    R(t) = [ 60t/t^2 +9 ] + 35

    Approx. how many barrels of oil will be produced in the first 5 years of production?!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by melanie View Post
    Here's a problem. Not sure if I should just sub in or firstly integrate and then sub in. Please help..

    an oil company found that the rate of oil production from an oil well is given by a function of time t, in years:

    R(t) = [ 60t/t^2 +9 ] + 35

    Approx. how many barrels of oil will be produced in the first 5 years of production?!
    you want \int_0^5R(t)~dt
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  3. #3
    Newbie melanie's Avatar
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    Thank you very much for that tip!!!

    But another question is . What happens to the constant (35) when integrating. DO i just ignore it?!

    So, (im not sure how to use the integration sign on the computer)

    'Integration sign (0-5 years) = |' [60t/t^2+9] +53 dt
    = 60/2 | [1/t^2+9] 2t dt

    let u = t^2 +9
    then du = 2t dt (?)

    = 60/2 | 1/u du

    = 60/2 ln |u| + c

    = 30 ln (t^2 +9 )

    = 30 ln 34 - 30 ln 9
    = 105.8 - 65.92
    = approx. 40 ?

    I really appreciate your help !!
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by melanie View Post
    Thank you very much for that tip!!!

    But another question is . What happens to the constant (35) when integrating. DO i just ignore it?!

    So, (im not sure how to use the integration sign on the computer)

    'Integration sign (0-5 years) = |' [60t/t^2+9] +53 dt
    = 60/2 | [1/t^2+9] 2t dt

    let u = t^2 +9
    then du = 2t dt (?)

    = 60/2 | 1/u du

    = 60/2 ln |u| + c

    = 30 ln (t^2 +9 )

    = 30 ln 34 - 30 ln 9
    = 105.8 - 65.92
    = approx. 40 ?

    I really appreciate your help !!
    the rule is, \int cx^n~dx = \frac {cx^{n + 1}}{n + 1} + C

    note that if you have a constant, the n above is zero. since c = cx^0

    thus, \int c~dx = cx + C

    where c is a constant
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  5. #5
    Newbie melanie's Avatar
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    Hmm.. ok.

    So, I integrate 60t/t^2+9 AND 35 separately.

    I get:

    30 ln 34 - 30 ln 9 = 215 [ for the first one ]

    35(5) - 35(0) = 175 [ from integrating the constant ]

    and I add them both together:
    = 215.

    Is that right ?! :S
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by melanie View Post
    Hmm.. ok.

    So, I integrate 60t/t^2+9 AND 35 separately.

    I get:

    30 ln 34 - 30 ln 9 = 215 [ for the first one ]

    35(5) - 35(0) = 175 [ from integrating the constant ]

    and I add them both together:
    = 215.

    Is that right ?! :S
    your final answer is right. but 30ln(34) - 30ln(9) is not 215

    of course, the answer is approximate, so you should write the squiggly equal sign
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  7. #7
    Newbie melanie's Avatar
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    oooops !! i meant, 39.87 for that one! EXCELLENT! thanks you soo much!!!
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