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# 数学代写|微积分代考calculus assignment|Covering Relations

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## 数学代写|微积分代考calculus assignment|First mean-value theorem for integrals

The original Newton integral, the student will recall, requires of indefinite integrals that the derivative requirement holds at every point (no exceptional set is allowed). Let us return to that briefly.
How can we determine the value of a definite integral
$$\int_{a}^{b} f(x) d x$$
for a function $f$ ? According to the definition we need to find an indefinite integral $F$ first and then compute $F(b)-F(a)$. Finding an indefinite integral may be a much harder task than simply evaluating this single number $F(b)-F(a)$.

The mean value theorem for derivatives gives a hint. According to that theorem
$$F(b)-F(a)=f(\xi)(b-a)$$
for at least one point $\xi$ in $(a, b)$. That gives the identity
$$\int_{a}^{b} f(x) d x=f(\xi)(b-a)$$
but we do not know precisely which point $\xi$ to choose. This result is called the first mean-value theorem for the integral; we see it is available for the narrowest version of the Newton integral, the one where the indefinite integral $F$ has the integrand $f$ as its derivative at every point inside the interval.

This relation between an interval $[a, b]$ and some selected point $\xi$ is called a covering relation. While the covering relation suggested by the first mean-value theorem for integrals is a useful one it cannot be made the basis for defining an integral.

## 数学代写|微积分代考calculus assignment|Riemann sums

The identity
$$\int_{a}^{b} f(x) d x=f(\xi)(b-a)$$
that we have just seen might be improved by subdividing the interval $[a, b]$ by intermediate points:
$$a=a_{1}<b_{1}=a_{2}<b_{2}=a_{3}<\cdots<a_{n}<b_{n}=b$$
This expresses the interval $[a, b]$ as the union of a collection of $n$ nonoverlapping, compact subintervals:
$$\left[a_{1}, b_{1}\right],\left[a_{2}, b_{2}\right],\left[a_{3}, b_{3}\right], \ldots,\left[a_{n}, b_{n}\right]$$
The mean-value theorem of the differential calculus, as before, asserts that we can select a point $\xi_{i}$ inside each interval $\left[a_{i}, b_{i}\right]$ so that
$$F\left(b_{i}\right)-F\left(a_{i}\right)=f\left(\xi_{i}\right)\left(b_{i}-a_{i}\right) .$$
This leads to a new covering relation
$$\pi=\left{\left(\left[a_{1}, b_{1}\right], \xi_{1}\right),\left(\left[a_{2}, b_{2}\right], \xi_{2}\right),\left(\left[a_{3}, b_{3}\right], \xi_{3}\right), \ldots,\left(\left[a_{n}, b_{n}\right], \xi_{n}\right}\right.$$
which is called a partition. The partition is denoted as $\pi$ (the letter is chosen so as to use the Greek letter corresponding to “P,” not to have anything to do with areas of circles).
Using this partition $\pi$ we have
$$F(b)-F(a)=\sum_{i=1}^{n}\left[F\left(b_{i}\right)-F\left(a_{i}\right)\right]=\sum_{i=1}^{n} f\left(\xi_{i}\right)\left(b_{i}-a_{i}\right)$$
and consequently
$$\int_{a}^{b} f(x) d x=\sum_{i=1}^{n} f\left(\xi_{i}\right)\left(b_{i}-a_{i}\right)$$
The sums here are called Riemann sums for the function $f$ (or sometimes, more correctly, Cauchy sums) and there is a long history of using such sums and partitions of intervals to estimate integrals.

Again, however, this discussion does not move us much closer to finding a definition of the integral since we would be unable to choose the correct partition to use unless we already knew F and f.

## 数学代写|微积分代考CALCULUS ASSIGNMENT|Full covers and Cousin covers

We prefer to say merely that $\beta$ is a full cover if $\beta$ is a full cover of $\mathbb{R}$, i.e., if $\beta$ is full at every real number. The entire theory of differentiation and integration of the calculus can be presented in a way that directly relates to full covers.

If a covering relation $\beta$ is a full cover then we have expressed the opinion that it should contain a partition of any interval $[a, b]$, i.e., there should be a subset $\pi$ of $\beta$,
$$\pi=\left{\left(\left[a_{i}, b_{i}\right], \xi_{i}\right): i=1,2, \ldots, n\right}$$
so that the intervals
$$\left{\left[a_{i}, b_{i}\right]: i=1,2, \ldots, n\right}$$
form a nonoverlapping collection of subintervals that make up all of $[a, b]$.
Note that, if our goal is to have partitions of $[a, b]$, we do not quite need $\beta$ to be full at the endpoints $a$ and $b$ since we would use only subintervals of $[a, b]$ and not concern ourselves with what is happening on the left at $a$ or what is happening on the right at $b$. This leads to the following definition, which slightly relaxes the condition defining full covers and also focusses on our need for partitions.

DEFINITION 2.3. A covering relation $\beta$ is a Cousin cover of the compact interval $[a, b]$ provided that, at each point $x$ in $[a, b]$, there is a $\delta>0$ so that $\beta$ contains all pairs $([c, d], x)$ for which $c \leq x \leq d,[c, d] \subset[a, b]$ and $0<d-c<\delta$.

## 数学代写|微积分代考CALCULUS ASSIGNMENT|FIRST MEAN-VALUE THEOREM FOR INTEGRALS

∫一种bF(X)dX

F(b)−F(一种)=F(X)(b−一种)

∫一种bF(X)dX=F(X)(b−一种)

## 数学代写|微积分代考CALCULUS ASSIGNMENT|RIEMANN SUMS

$$\int_{a}^{b} f(x) d x=f(\xi)(b-a)$$
that we have just seen might be improved by subdividing the interval $[a, b]$ by intermediate points:
$$a=a_{1}<b_{1}=a_{2}<b_{2}=a_{3}<\cdots<a_{n}<b_{n}=b$$
This expresses the interval $[a, b]$ as the union of a collection of $n$ nonoverlapping, compact subintervals:
$$\left[a_{1}, b_{1}\right],\left[a_{2}, b_{2}\right],\left[a_{3}, b_{3}\right], \ldots,\left[a_{n}, b_{n}\right]$$
The mean-value theorem of the differential calculus, as before, asserts that we can select a point $\xi_{i}$ inside each interval $\left[a_{i}, b_{i}\right]$ so that
$$F\left(b_{i}\right)-F\left(a_{i}\right)=f\left(\xi_{i}\right)\left(b_{i}-a_{i}\right) .$$
This leads to a new covering relation
$$\pi=\left{\left(\left[a_{1}, b_{1}\right], \xi_{1}\right),\left(\left[a_{2}, b_{2}\right], \xi_{2}\right),\left(\left[a_{3}, b_{3}\right], \xi_{3}\right), \ldots,\left(\left[a_{n}, b_{n}\right], \xi_{n}\right}\right.$$
which is called a partition. The partition is denoted as $\pi$ (the letter is chosen so as to use the Greek letter corresponding to “P,” not to have anything to do with areas of circles).
Using this partition $\pi$ we have
$$F(b)-F(a)=\sum_{i=1}^{n}\left[F\left(b_{i}\right)-F\left(a_{i}\right)\right]=\sum_{i=1}^{n} f\left(\xi_{i}\right)\left(b_{i}-a_{i}\right)$$
and consequently
$$\int_{a}^{b} f(x) d x=\sum_{i=1}^{n} f\left(\xi_{i}\right)\left(b_{i}-a_{i}\right)$$

## 数学代写|微积分代考CALCULUS ASSIGNMENT|FULL COVERS AND COUSIN COVERS

$$\pi=\left{\left(\left[a_{i}, b_{i}\right], \xi_{i}\right): i=1,2, \ldots, n\right}$$
so that the intervals
$$\left{\left[a_{i}, b_{i}\right]: i=1,2, \ldots, n\right}$$

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