Given a simple closed plane curve, find a relationship between the area of the original curve and the area of its parallel curve For context, I have been introduced to simple plane curves from the perspective of single-variable calculus through the use of parameterization. Furthermore, my textbook only deals with plane curves that don't intersect themselves.
The question from my textbook is stated as follows:

Let $x = x(t)$, $y = y(t)$ be a closed curve. A constant length $d$ is measured off along the normal to the curve. The extremity of this segment describes a curve which is called a parallel curve to the original curve. Find the area, the length of arc, and the radius of curvature of the parallel curve

Now, after much rumination, I have found expressions for the length of arc and radius of curvature of the parallel curve. The only one I have been having a lot of trouble with is the area of the parallel curve.
The first approach that came to mind was a simple algebraic approach:

Since the question did not specify, I just assumed that the curve is arc-length parameterized to make life a little easier. Then, the parallel curve can be represented by the following parametric equations:
$$x_p=\dot x - R\ddot y, \\ y_p = \dot y + R\ddot x$$
Now, the area of a parametric curve is normally given by the equation:
$$A = \int^{t_1}_{t_0}y\dot xdt$$
We can obtain an alternative expression for the area using integration by parts:
$$\begin{align}A &= y(t_1)x(t_1)-y(t_0)x(t_0)-\int^{t_1}_{t_0}x\dot ydt \\ &= -\int^{t_1}_{t_0}x\dot ydt, \text{since the curve is closed } x(t_1) = x(t_0),y(t_1) = y(t_0) \end{align}$$
Combining both expressions above, we obtain a nice symmetrical form for the equation for area:
$$A=\frac{1}{2}\int^{t_1}_{t_0}y\dot x - x\dot y dt$$
Now, if we denote the total length of the original curve by $l$, the area of the parallel curve is:
$$\begin{align} A_p &= \frac{1}{2}\int^{l}_{0}(y-R\dot x)(\dot x + R\ddot y)-(x+R\dot y)(\dot y - R\ddot x)dt \\ &= \frac{1}{2}\int^{l}_{0} (y\dot x - x\dot y) + R(y\ddot y + x\ddot x) - R(\dot x^2 + \dot y^2) - R^2(\dot x \ddot y - \dot y \ddot x)dt \\ &= \frac{1}{2}\int^{l}_{0} (y\dot x - x\dot y) + R(y(k\dot x) + R(-k\dot y)) - R - R^2k dt, \text{ where } k \text{ refers to the curvature of the original curve} \\ &= \frac{1}{2}\int^{l}_{0} (y\dot x - x\dot y)(1+Rk)-R-R^2k dt \end{align}$$

I didn't really know how to continue simplifying this integral any further from this point, so I tried to get a better intuitive grasp of the problem by considering the simplest case of a plane curve - the circle:

If you have a circle of radius $r$ and offset this circle by a distance $d$, the relationship between the area of the offset circle and the original circle can be stated as:
$$\begin{align} \pi (r+d)^2 &= \pi r^2 + 2\pi rd + \pi d^2 \\ A_p &= A + Cd + \pi d^2, \text{ where } C \text{ refers to the circumference of original circle} \end{align}$$

The equation above was interesting because it is similar in form to the integrand I obtained before. The only way I could think of to make use of this simpler case was to consider the osculating circles of the original curve and the parallel curve; however, I wasn't successful in fleshing out this idea further.
I would appreciate any input on how I could either flesh out my idea of using osculating circles or any hints on simplifying the integral I got. Any other ideas on how to approach this problem are welcome as well.
 A: Let's assume that the curve $\alpha(t) = (x,y) \colon [0,l] \rightarrow \mathbb{R}^2$ is a simple, closed curve parametrized by arc-length and oriented counterclockwise. Since there are two possible normals, let's assume that we want to move in the direction of the normal which points "outside" the curve. This means that the new curve $\beta$ is given by
$$ \beta(t) = \alpha(t) - d N(t) = (x(t),y(t)) - d(N_1(t),N_2(t)) = (x(t) - d N_1(t), y(t) - dN_2(t)). $$
Then by the Frenet-Serret formula for plane curves, we have
$$ \dot{\beta}(t) = \dot{\alpha}(t) - d \dot{N}(t) = (1 + d k(t)) \cdot \dot{\alpha}(t) = (1 + d k(t)) \cdot (\dot{x}(t),\dot{y}(t)). $$
By Green's formula for the area enclosed by $\beta$, we have
$$ \operatorname{Area}(\beta) = \frac{1}{2} \int_0^l (1 + d k(t)) \left( \left( y(t) - d N_2(t) \right) \dot{x}(t) - \left( x(t) - d N_1(t) \right) \dot{y}(t) \right) \, dt = \\
\operatorname{Area}(\alpha) +
\underbrace{\frac{d}{2} \int_0^l k(t) \left( y(t) \dot{x}(t) - x(t) \dot{y}(t) \right) \, dt }_{(1)} +
\underbrace{\frac{d}{2} \int_0^l N_1(t) \dot{y}(t) - N_2(t) \dot{x}(t) \, dt}_{(2)} +
\underbrace{\frac{d^2}{2} \int_0^l k(t) \cdot \left(  N_1(t) \dot{y}(t) - N_2(t) \dot{x}(t) \right) \, dt}_{(3)}. $$
Since $\dot{N}(t) = -k(t) \dot{\alpha}(t)$, we can write
$$ k(t) y(t) \dot{x}(t) - k(t) x(t) \dot{y}(t) = \dot{N_2}(t) x(t) - \dot{N_1}(t) y(t). $$
Applying integration by parts, we get that
$$ (1) = \frac{d}{2} \int_0^l \dot{N_2}(t) x(t) - \dot{N_1}(t) y(t) \, dt = \frac{d}{2} \int_0^l N_1(t) \dot{y}(t) - N_2(t) \dot{x}(t) \, dt = (2). $$
Since $\left( \left(\dot{x}(t),\dot{y}(t) \right), \left( N_1(t),N_2(t) \right)\right)$ is positive-oriented orthonormal frame, this means that if we rotate $(\dot{x}(t),\dot{y}(t))$ by 90 degrees counterclockwise, we get $\left( \dot{y}(t), -\dot{x}(t) \right) = \left( N_1(t), N_2(t) \right)$ and so
$$ N_1(t) \dot{y}(t) - N_2(t) \dot{x}(t) = \| N(t) \|^2 \equiv 1. $$
Hence,
$$ (1) + (2) = d \int_0^l N_1(t) \dot{y}(t) - N_2(t) \dot{x}(t) \, dt = d \int_0^l 1 \, dt = ld $$
while by the total curvature theorem, we have
$$ (3) = \frac{d^2}{2} \int_0^l k(t) \, dt =  \frac{d^2}{2} \cdot (2\pi) = \pi d^2. $$
Combining everything, we get that
$$ \operatorname{Area}(\beta) = \operatorname{Area}(\alpha) + ld + \pi d^2 $$
as expected.
