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Calculus problem

  1. Mar 1, 2016 #1
    1. The problem statement, all variables and given/known data given## dm/dt=120(t-3)^2## find m(t)


    2. Relevant equations


    3. The attempt at a solution
    now i have a problem here the first attempt ## ∫120(t-3)^2dt ##
    = ##120∫(t^2-6t+9)dt##
    =## 120(t^3/3-3t^2+9t)+k ##
    but if you use chain rule you have
    ## ∫120(t-3)^2dt ## let ## u = t-3 ##then it follows that ##du=dt## thus we have
    ##∫120u^2du## =##(120u^3)/3 + k ##=##(120(t-3)^3)/3 ## this second solution clearly is the correct answer having four terms + a constant, why is the first solution wrong ? its having three terms + a constant?
     
  2. jcsd
  3. Mar 1, 2016 #2

    Samy_A

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    Both seem correct to me. The fourth term in your second solution is just a number, so you can absorb it in the integration constant.
     
  4. Mar 1, 2016 #3

    Ssnow

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    The solution are the same if in the second method you call ##-9\cdot 120 + k = c## a constant, both solution are of the same form ##120(t^3/3-3t^{2}+9t) +k## and both differ only by a constant term...
     
  5. Mar 1, 2016 #4
    can one take the first solution as the answer to the differential equation?
     
  6. Mar 1, 2016 #5

    Samy_A

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    Yes
     
  7. Mar 1, 2016 #6
    Thanks Sam and snow. regards
     
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