# Integrals problem

1. Jun 9, 2016

### Nanu Nana

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. Jun 9, 2016

### Delta²

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

3. Jun 9, 2016

### Nanu Nana

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

4. Jun 9, 2016

### Delta²

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

5. Jun 9, 2016

### Nanu Nana

How do we get that -3y ???

6. Jun 9, 2016

### Delta²

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. Jun 9, 2016

### Nanu Nana

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

8. Jun 9, 2016

### 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).