Find a scalar potential φ for the vector field F such that $\varphi(0,0)=5$ when
$$\mathbf{F}(x,y) = {3\cdot x\cdot \left(2\cdot y+1\right)}\mathbf{i}+{3\cdot x^2}\mathbf{j}$$
First we need to check that vector field is conservative so
$$\frac{\delta F_1}{\delta y}=\frac{\delta F_2}{\delta x}$$
$$6x=6x\to \frac{\delta F_1}{\delta y}-\frac{\delta F_2}{\delta x} = 0$$
then I want to find the a scalar potential $\varphi$ for the vector field $\mathbf F$ such that $\varphi(0,0)=5$.
So
$$f_x ={3\cdot x\cdot \left(2\cdot y+1\right)} \to f= 3\cdot x^2 \cdot y^2+\frac{3\cdot x^2}{2}+g(y) $$
then
$$f_y=3\cdot x^2+g'(y)=3\cdot x^2\to g(y)=k$$
and $$f(x,y)= 3\cdot x^2 \cdot y^2+\frac{3\cdot x^2}{2}+k$$
and $\varphi$ for the vector field $\mathbf F$ such that $\varphi(0,0)=5$.
$$3\cdot 0^2 \cdot 0^2+\frac{3\cdot 0^2}{2}+k=5 \to k=5$$ but the $k=5$ is wrong answer. I have done some stupid mistake but I can't see where.
We have that
$$f_x ={3\cdot x\cdot \left(2\cdot y+1\right)} \implies f= 3\cdot x^2 \cdot y+\frac{3\cdot x^2}{2}+g(y)$$
$$f_y=3\cdot x^2+g'(y)=3\cdot x^2\implies g(y)=k$$
and therefore
$$f(x,y)= 3\cdot x^2 \cdot y+\frac{3\cdot x^2}{2}+k$$
but in any case $k=5$ for the given condition.