Integrals problem

  • Thread starter Nanu Nana
  • Start date
  • #1
95
5

Homework Statement



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

Homework Equations


y(x)=∫ x^2 dx




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 ??
 

Answers and Replies

  • #2
Delta2
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Nope you just didn't notice that C=3,333....=10/3.
 
  • #3
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But then shouldn't it be x3 +10/3 ??
 
  • #5
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How do we get that -3y ???
 
  • #6
Delta2
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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##??
 
  • #7
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Yeah I couldn't see it but now I do . Thanks
 
  • #8
Delta2
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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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