Answer
a) $f(x,y)=\frac{1}{3}x^3y^3$
b) $\int_{C}\textbf{F}\cdot d \textbf{r}=-9$
Work Step by Step
We are given:
$$\textbf{F}(x,y)=x^2y^3\: \textbf{i} \: + x^3y^2\: \textbf{j}$$
First, let us integrate the $\textbf{i}$ component of the vector-valued function with respect to $x$.
$$f(x,y)=\int x^2y^3 dx = \frac{1}{3} x^3y^3 + g(y)$$
Note that the constant of integration is denoted as $g(y)$ since the resultant function is a constant with respect to $x$.
Taking the derivative of the function with respect to $y$, we should get the $\textbf{j}$ component of the vector-valued function.
$$f_{y} (x,y) = x^3y^2 + g'(y)$$
Comparing with our original vector-valued function, we can deduce: $$g(y)=0\therefore \\f(x,y)=\frac{1}{3}x^3y^3$$
The fundamental theorem of line integrals states:
$$\int_{C}\nabla\textbf f \cdot d\textbf{r} = f(b)-f(a)$$
We can work out the points $a$ and $b$ using the parametrization of the curve.
At $t=0$, the parametric curve is at $(0^3-2(0), 0^3+2(0))=(0,0)$
At $t=1$, the parametric curve is at $(1^3-2(1),1^3+2(1))=(-1,3)$
Using this, we can evaluate the line integral.
$$\int_{C} \langle x^2 y^3, x^3 y^2 \rangle \cdot d \textbf{r}=f(-1,3)-f(0,0) \\
=\frac{1}{3}(-1)^\: 3^3-\frac{1}{3}0^3\:0^3\\
=-9$$
