# 微积分作业代写calclulus代考| Multiple integrals

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## 微积分作业代写calclulus代考|Integration of f : I ⊂ R −→ R

Suppose $f$ is a continuous function of $x$ and assume that the interval $I$ is closed and bounded and lying in the function domain, $D_{f}$. That is,
$$I={x: a \leq x \leq b} \subset D_{f} .$$
The graph of $f$ is a curve in $\mathbb{R} \times \mathbb{R}=\mathbb{R}^{2}$ as shown in Figure $4.1$ below.

First the interval $I$ is cut $I$ into small bits – this is called a partition of $I$ :
$$a=x_{0}<x_{1}<x_{2}<\ldots<x_{n-1}<x_{n}=b$$
with $n$ subintervals $I_{i}=\left{x: x_{i-1} \leq x \leq x_{i}\right}$, of width $\Delta x_{i}=x_{i}-x_{i-1}$. A few of these subintervals are shown in Figure 4.1.

Then, choosing some real number $\xi_{i} \in I_{i}$ from each subinterval, we form the sum
$$\sigma_{n}=\sum_{i=1}^{n} f\left(\xi_{i}\right) \Delta x_{i}$$
This it called the Riemann sum of $f$ over $I$. From its construction we see that it must depend on the partition of $n$ subintervals.

## 微积分作业代写calclulus代考|The geometric interpretation of σn for f : I −→ R

If $f \geq 0$, then $f\left(\xi_{i}\right) \Delta x_{i}$ is the area of the rectangle of height $f\left(\xi_{i}\right)$ and width $\Delta x_{i}$ as shown here in Figure 4.2.
Hence, the sum $\sigma_{n}$ is an approximation to the area “under” the curve $y=f(x)$ and over $I$.

To improve on this approximation we find numbers $\ell_{i}$ and $m_{i}$ in each interval $I_{i}$ such that $f\left(\ell_{i}\right) \leq f(x) \leq f\left(m_{i}\right)$ for all $x$ in $I_{i}$.
For a given partition we form the upper and lower sums
$$R_{\min }=\sum_{i=1}^{n} f\left(\ell_{i}\right) \Delta x_{i} \leq \sum_{i=1}^{n} f\left(\xi_{i}\right) \Delta x_{i} \leq \sum_{i=1}^{n} f\left(m_{i}\right) \Delta x_{i}=R_{\max }$$
In this process we have constructed upper and lower bounds on $\sigma_{n}$. That is,
$$R_{\min } \leq \sigma_{n} \leq R_{\max }$$
We now take the simultaneous limit of the number of intervals $n \rightarrow \infty$ and the representative size of the intervals $\max \left(\Delta x_{i}\right) \rightarrow 0$. We find that, as $n \rightarrow \infty, R_{\min }$ increases and $R_{\max }$ decreases. If the dual limits exist and $\lim R_{\min }=\lim R_{\max }$, then an application of a squeeze theorem gives:

## 微积分作业代写calclulus代考|Integration of f : I ⊂ R −→ R

σn=∑一世=1nF(X一世)ΔX一世

## 微积分作业代写calclulus代考|The geometric interpretation of σn for f : I −→ R

R分钟=∑一世=1nF(ℓ一世)ΔX一世≤∑一世=1nF(X一世)ΔX一世≤∑一世=1nF(米一世)ΔX一世=R最大限度

R分钟≤σn≤R最大限度