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Integrals problem

  1. Jun 9, 2016 #1
    1. The problem statement, all variables and given/known data

    Determine the function for which y '= x ^ 2, and (2,6) is a point of the graph.
    2. Relevant equations
    y(x)=∫ x^2 dx




    3. The attempt at a solution
    I tried doing this but didn't get the right answer
    y(x)=∫ x^2 dx
    = x^3/3 +c

    x=2 y=6

    6=(2)^3 /3 +c
    C= 3,33...
    But according to the correctionpaper, its wrong . Answer should be x^3-3y+10 =0

    How ??
     
  2. jcsd
  3. Jun 9, 2016 #2
    Nope you just didn't notice that C=3,333....=10/3.
     
  4. Jun 9, 2016 #3
    But then shouldn't it be x3 +10/3 ??
     
  5. Jun 9, 2016 #4
    ##x^3/3 + 10/3=y##
     
  6. Jun 9, 2016 #5
    How do we get that -3y ???
     
  7. Jun 9, 2016 #6
    So you mean you did the integral and finding C by yourself but you cant see the algebraic equivalence of ##(x^3+10)/3=y## and ##x^3+10-3y=0##??
     
  8. Jun 9, 2016 #7
    Yeah I couldn't see it but now I do . Thanks
     
  9. Jun 9, 2016 #8
    Ok you welcome, maybe you are not used in the perplexed form involving x and y, you are used in the form y=f(x).
     
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