Solve Integral: y'=x^2, Point (2,6) on Graph

[Nanu Nana]

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 ??

 

[Delta2]

Nope you just didn't notice that C=3,333...=10/3.

 

[Nanu Nana]

But then shouldn't it be x3 +10/3 ??

 

[Delta2]

$x^3/3 + 10/3=y$

 

[Nanu Nana]

How do we get that -3y ?

 

[Delta2]

So you mean you did the integral and finding C by yourself but you can't see the algebraic equivalence of $(x^3+10)/3=y$ and $x^3+10-3y=0$??

 

[Nanu Nana]

Yeah I couldn't see it but now I do . Thanks

 

[Delta2]

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).