# Proof of Fundamental Theorem for line integrals

## Description

I've been looking into the proof of Fundamental Theorem for line integrals in Multivariable Calculus course officially provided by MIT(18.02sc).
And the question is that at the final step, I could not follow how the transition in the proof below happening at the part below $$\int^{t_1}_{t_0} \frac{d}{dt} f(x(t), y(t)) dt = f(x(t), y(t)) \rvert^{t_1}_{t_0}$$

## Theorem Fundamental Theorem for line integrals

If $$\mathbf{F} = \mathbf{\nabla} f$$ is a gradient field of $$f$$ and $$C$$ is any curve with endpoints $$P_0 = (x_0, y_0)$$ and $$P_1 = (x_1, y_1)$$ then $$\int_C \mathbf{F} \cdot d \mathbf{r} = f(x, y) \rvert^{P_1}_{P_0} = f(x_1, y_1) - f(x_0, y_0)$$

## Proof: Theorem Fundamental Theorem for line integrals

$$\int_C \mathbf{F} \cdot d \mathbf{r} = \int_C f_x dx + f_y dy = \int^{t_1}_{t_0} \Big[ f_x(x(t), y(t)) \frac{dx}{dt} + f_y(x(t), y(t)) \frac{dy}{dt} \Big] dt = \int^{t_1}_{t_0} \Big[ \frac{d}{dx} f(x(t), y(t)) \frac{dx}{dt} + \frac{d}{dy} f(x(t), y(t)) \frac{dy}{dt} \Big] dt = \int^{t_1}_{t_0} \frac{d}{dt} f(x(t), y(t)) dt = f(x(t), y(t)) \rvert^{t_1}_{t_0} = f(P_1) - f(P_0)$$

• Well actually you might be overthinking it as integration is the reverse operation of differentiation so $\int ... dt$ and the $\frac{d}{dx}$ cancel out. Commented May 22, 2020 at 8:26

Let $$F(t):= f(x(t), y(t)),$$ then
$$\int^{t_1}_{t_0} \frac{d}{dt} f(x(t), y(t)) dt= \int^{t_1}_{t_0}F'(t) dt = F(t_1)-F(t_0).$$
• oh,,, sorry. I was being careless and too much focused on the point first term($\frac{d}{dt}$) and the last $dt$... Thank you for your answer! Commented May 22, 2020 at 8:29