微积分网课代修|积分学代写Integral Calculus代考|MAT272 The General Newton integral

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微积分作业代写calclulus代考|The General Newton integral

The version of the Newton integral promoted in this chapter can be described this loose way:
We write
$$\int_{a}^{b} f(x) d x=F(b)-F(a)$$
provided that we can determine a continuous function $F:[a, b] \rightarrow \mathbb{R}$ so that $F^{\prime}(x)=f(x)$ at all but a negligible set of points in $(a, b)$.
By negligible here we mean the set of points $X$ where the identity $F^{\prime}(x)=f(x)$ might fail or simply be unknown can be written as a sequence. The justification, again in loose language is this:
a continuous function $F:[a, b] \rightarrow \mathbb{R}$ cannot grow on a negligible set of points $e_{1}, e_{2}, e_{3}, \ldots$
By the early twentieth century it was recognized that a larger class of negligible sets was needed for many problems. The class of sets that we might need to neglect are called the null sets. A null set is small in certain senses, but might be too large to be written out as a sequence of points. It turns out that while continuous functions do not grow on sequences of points, they might grow on null sets. Thus the definition of the Newton integral requires both a relaxation in the class of sets to be neglected and a tightening of the requirement on the function $F$. The definition of the modern version of the Newton integral is this:
We write
$$\int_{a}^{b} f(x) d x=F(b)-F(a)$$
meaning that there is a continuous function $F:[a, b] \rightarrow \mathbb{R}$ so that:
(a) There is a null set $N$,
(b) $F^{\prime}(x)=f(x)$ for all $x$ in $(a, b)$ except possibly for $x$ in the null set $N$, and
(c) this function $F$ does not grow on the negligible set $N$.
This integral is properly defined as a Newton integral in Chapter 3 . An equivalent constructive definition is given in Chapter 4. A measure-theoretic account is given in Chapter 5 . This is the correct integral for the calculus.

微积分作业代写calclulus代考|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.

微积分作业代写calclulus代考|The General Newton integral

$$\int_{a}^{b} f(x) d x=F(b)-F(a)$$

$$\int_{a}^{b} f(x) d x=F(b)-F(a)$$

(a) 有一个空集 $N$ ，
(b) $F^{\prime}(x)=f(x)$ 对所有人 $x$ 在 $(a, b)$ 除了可能 $x$ 在零集中 $N$, 和
(c) 这个函数 $F$ 在可忽略的集合上不增长 $N$.

微积分作业代写calclulus代考|First mean-value theorem for integrals

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$$\int_{a}^{b} f(x) d x$$

$$F(b)-F(a)=f(\xi)(b-a)$$

$$\int_{a}^{b} f(x) d x=f(\xi)(b-a)$$