# 微积分网课代修|积分学代写Integral Calculus代考|MA1030C Larger exceptional sets

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## 微积分作业代写calclulus代考|Larger exceptional sets

As integration theory developed over the centuries since Newton it became clear that the theory required quite large exceptional sets, certainly larger than just a few points. But this also requires us to characterize those sets that can be so neglected and also to describe what we must require of an indefinite integral so that these sets can be ignored.

At a calculus level we can easily go one step further, even if we cannot quite approach the full generality needed. The key is to push Lemma $1.2$ and Lemma $1.3$ much further. We cannot do this with help from the mean-value theorem as before: indeed the proof is deferred to the next chapter where we introduce a new technique needed for the integration theory.

LEMMA 1.5. Suppose that $F$ and $G$ are both continuous functions on an interval $[a, b]$. Suppose that there is a sequence of points $e_{1}, e_{2}, e_{3}, \ldots$ of points from $[a, b]$ and that $F^{\prime}(x)=G^{\prime}(x)$ for all $a<x<b$ except possibly at the points $e_{1}, e_{2}, e_{3}$, … Then
$$F(b)-F(a)=G(b)-G(a)$$

## 微积分作业代写calclulus代考|A version of the Newton integral for elementary calculus

Assuming Lemma $1.5$ for the moment we can introduce a much improved version of the Newton integral.

Definition $1.6$ (Modified Newton Integral). Suppose that $f$ is a function defined on an interval $(a, b)$ except possibly at the points of a sequence $e_{1}, e_{2}, e_{3}$, .. from $[a, b]$. Suppose that we can find a continuous function $F:[a, b] \rightarrow \mathbb{R}$ so that $F^{\prime}(x)=f(x)$ for every $x$ with $a<x<b$ with perhaps the exception of the points $e_{1}, e_{2}, e_{3}, \ldots$. Then we will say that $F$ is an indefinite integral of $f$ on $[a, b]$ and we will write
$$\int_{a}^{b} f(x) d x=F(b)-F(a)$$
and call the latter the definite integral of $f$ on $[a, b]$.
The only justification needed would be to use Lemma $1.5$ to check that if $F$ and $G$ both qualify to be indefinite integrals of $f$ on an interval $[a, b]$, then $F$ and $G$ differ by a constant so that $F(b)-F(a)=G(b)-G(a)$. Thus the definite integral is unambiguously defined.

There is one subtle point here that might be missed. Suppose that two functions $F$ and $G$ both qualify to be indefinite integrals of $f$. That means that there is some sequence of points $e_{1}, e_{2}, e_{3}, \ldots$ from $[a, b]$ and that $F^{\prime}(x)=f(x)$ provided $x$ is not one of the points in this sequence. It also means that there is some sequence of points $e_{1}^{\prime}, e_{2}^{\prime}, e_{3}^{\prime}, \ldots$ from $[a, b]$ [not necessarily the same sequence as before] and that $G^{\prime}(x)=f(x)$ provided $x$ is not one of the points in this sequence.

Accordingly, we observe that $F^{\prime}(x)=G^{\prime}(x)$ except possibly at points belonging to the combined sequence
$$e_{1}, e_{1}^{\prime}, e_{2}, e_{2}^{\prime}, e_{3}, e_{3}^{\prime}, e_{4}, e_{4}^{\prime}, \ldots$$allowing us to apply Lemma 1.5. From that lemma we deduce that $F$ and $G$ differ by a constant and that $F(b)-F(a)=G(b)-G(a)$. Thus the definite integral is unambiguously defined no matter what indefinite integral we choose to use.

## 微积分作业代写calclulus代考|Larger exceptional sets

$$F(b)-F(a)=G(b)-G(a)$$

## 微积分作业代写calclulus代考|A version of the Newton integral for elementary calculus

$F^{\prime}(x)=f(x)$ 对于每个 $x$ 和 $a<x<b$ 也许除了点 $e_{1}, e_{2}, e_{3}, \ldots$. 然后我们会说 $F$ 是 一个不定积分 $f$ 上 $[a, b]$ 我们会写
$$\int_{a}^{b} f(x) d x=F(b)-F(a)$$

$$e_{1}, e_{1}^{\prime}, e_{2}, e_{2}^{\prime}, e_{3}, e_{3}^{\prime}, e_{4}, e_{4}^{\prime}, \ldots$$