这是一份非参数统计作业代写的成功案

For each $i$ compute $f\left(\Psi\left(t_{i}\right)\right) \delta t_{i}$. A little algebraic trickery yields
$$
\begin{aligned}
f\left(\Psi\left(t_{i}\right)\right) \delta t_{i} &=f\left(\Psi\left(t_{i}\right)\right) \delta t_{i} \frac{\Delta t_{i}}{\Delta t_{i}} \
&=f\left(\Psi\left(t_{i}\right)\right) \frac{\delta t_{i}}{\Delta t_{i}} \Delta t_{i}
\end{aligned}
$$
Sum over all $i$.
The integral $\int_{C} f(x, y) d \mathbf{s}$ is defined to be the limit of the sum from the previous step as $n \rightarrow \infty$. But as $n \rightarrow \infty$ it also follows that $\Delta t_{i} \rightarrow 0$. Our integrand contains the term $\frac{\delta_{i}}{\Delta t_{i}}$. As $\Delta t_{i} \rightarrow \infty$ this converges to $\left|\frac{\partial \Psi}{\partial t}\right|$. Hence, our integral has become
$$
\int_{C} f(x, y) d \mathbf{s}=\int_{a}^{b} f(\Psi(t))\left|\frac{\partial \Psi}{\partial t}\right| d t
$$

MAST90123/STAT 550/Math 776 COURSE NOTES ：

We now compute the area of each parallelogram.
Observe that each parallelogram is spanned by the vectors
$$
V_{u}=\Psi\left(u_{i+1}, v_{j}\right)-\Psi\left(u_{i}, v_{j}\right)
$$
and
$$
V_{v}=\Psi\left(u_{i}, v_{j+1}\right)-\Psi\left(u_{i}, v_{j}\right)
$$
The desired area is thus the magnitude of the cross product of these vectors:
$$
\text { Area }=\left|V_{u} \times V_{v}\right|
$$
We now do some algebraic tricks:
$$
\begin{aligned}
\left|V_{u} \times V_{v}\right| &=\left|V_{u} \times V_{v}\right| \frac{\Delta u \Delta v}{\Delta u \Delta v} \
&=\left|\frac{V_{u}}{\Delta u} \times \frac{V_{v}}{\Delta v}\right| \Delta u \Delta v
\end{aligned}
$$