Using the Fundamental Theorem of Line Integrals, evaluate $\int_{C} e^x \, dy + e^{x}y \, dx,$ where $C$ is the parabola $r(t)=\langle t+1,t^2 \rangle$ for $t\in[-1,3].$
What I tried is as follows. First, I used
$$\int_{C}e^{x}dy+e^{x}ydx=\int_{C}d(e^xy)=\int^{3}_{-1}d(e^xy).$$
Put $x=t+1, y=t^2$. Then, the line integral is
$$=\int^{3}_{-1}\frac{d}{dt}\bigg(e^{t+1}t^2\bigg)dt=e^{t+1}t^2\bigg|^{3}_{-1}.$$
I could also use that $x=t+1$ and $y=t^2$ to find that $y=(x-1)^2$ and $dy=2(x-1) \,dx.$ So, the line integral is $$\int^{4}_{0}2e^{x}(x-1)dx+e^x(x-1)^2dx.$$
Can anyone please tell me if my process is right? If not, then how do I solve it? Thanks.
Consider a two-dimensional vector field $\mathbf F(x, y) = \langle P(x, y), Q(x, y) \rangle$ and a curve $\mathcal C$ parametrized by the vector-valued function $\mathbf r(t) = \langle x(t), y(t) \rangle$ over the closed interval $[a, b].$ Given that there exists a twice continuously differentiable scalar function $f(x, y)$ on $\mathcal C$ such that $\nabla f(x, y) = \mathbf F(x, y),$ the Fundamental Theorem of Line Integrals states that $$\int_\mathcal C (P \, dx + Q \, dy) = \int_\mathcal C \mathbf F \cdot d \mathbf r = f(\mathbf r(b)) - f(\mathbf r(a)).$$ Can you find a function $f(x, y)$ such that $f_x(x, y) = e^x y$ and $f_y(x, y) = e^x$ that is twice continuously differentiable on the closed interval $[-1, 3]?$ If so, then the line integral is $f(4, 9) - f(0, 1)$ since $\mathbf r(-1) = \langle -1 + 1, (-1)^2 \rangle = \langle 0, 1 \rangle$ and $\mathbf r(3) = \langle 3 + 1, 3^2 \rangle = \langle 4, 9 \rangle.$