# 微积分网课代修|积分学代写Integral Calculus代考|MTH252 Integration formulas

• 单变量微积分
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## 微积分作业代写calclulus代考|Integration formulas

The Newton integral inherits its formulas directly from the standard differentiation formulas. If we review the latter we will be able to deduce useful and attractive formulas for the integral.

For the integral of Chapter 3 , which is much more general than the Newton versions here, these formulas remain true but will require some attention to hypotheses; they will not usually follow trivially from the differentiation formulas.
1.4.1. Sum formula. One of the first formulas we encounter in the calculus is that for the sum of two derivatives:
$$\frac{d}{d x}{F(x)+G(x)}=\frac{d}{d x} F(x)+\frac{d}{d x} G(x) .$$
From that we obtain the sum formula for integrals:
$$\int_{a}^{b}{f(x)+g(x)} d x=\int_{a}^{b} f(x) d x+\int_{a}^{b} g(x) d x .$$
The hypotheses allowing this are: $f$ has an indefinite integral $F$ and $g$ has an indefinite integral $G$ where both $F$ and $G$ are continuous on $[a, b]$ with
$$F^{\prime}(x)=f(x) \text { and } G^{\prime}(x)=g(x)$$
for all points in $(a, b)$ excepting possibly some sequence of exceptional points. This sum formula will be available for the Chapter 3 integral under much weaker hypotheses.
1.4.2. Integration by parts. One of the most studied of the formulas we encounter in the calculus is that for the product of two derivatives:
$$\frac{d}{d x}{F(x) G(x)}=F^{\prime}(x) G(x)+F(x) G^{\prime}(x)$$
From that we obtain the formula for integrals known as integration by parts:
$$\int_{a}^{b}{f(x) G(x)+F(x) g(x)} d x=F(b) G(b)-F(a) G(b) .$$
The hypotheses allowing this are: $f$ has an indefinite integral $F$ and $g$ has an indefinite integral $G$ where both $F$ and $G$ are continuous on $[a, b]$ with
$$F^{\prime}(x)=f(x) \text { and } G^{\prime}(x)=g(x)$$
for all points in $(a, b)$ excepting possibly some sequence of exceptional points. There are versions of integration by parts formulas for the general integration theory, but they require very different proofs.

## 微积分作业代写calclulus代考|Preview

The Newton integral is a sufficient tool for most of elementary calculus needs; we should be informed of some of its theory. One of the defects in the presentation to this point is that we do not know what functions can be integrated by this method.

To be sure if a given function $f$ is the derivative of some other function $F$ then we know how the procedure works. But what sufficient conditions can be stated for a function $f$ in order that we can be assured that such an indefinite integral exists? We cannot always be placed in the uncomfortable position of computing an indefinite integral in order to be assured that there is one.

We report here, by way of a preview, some of the theory that will clarify the situation. Proper statements of these facts appear in Chapter $3 .$
THEOREM 1.7. In order that a function $f$ possess an integral
$$\int_{a}^{b} f(x) d x$$
on a compact interval $[a, b]$ in the sense of the Newton integral of this chapter the following are sufficient:

• $f$ is continuous at every point of $[a, b]$.
• $f$ is continuous at every point of $(a, b)$ and is bounded.
• $f$ is continuous at every point of $(a, b)$ with the exception possibly of some sequence of points and $f$ is bounded.
• $f$ is continuous at every point of $(a, b)$ with at most finitely many exceptions and dominated by another function $g$ for which
$$0 \leq f(x) \leq g(x) \quad(a<x<b)$$
where $g$ is continuous on $(a, b)$ (again allowing finitely many exceptions) and the integral $\int_{a}^{b} g(x) d x$ is assumed to exist.

## 微积分作业代写calclulus代考|Integration formulas

1.4.1。求和公式。我们在微积分中遇到的第一个公式是两个导数之和:
$$\frac{d}{d x} F(x)+G(x)=\frac{d}{d x} F(x)+\frac{d}{d x} G(x) .$$

$$\int_{a}^{b} f(x)+g(x) d x=\int_{a}^{b} f(x) d x+\int_{a}^{b} g(x) d x .$$

$$F^{\prime}(x)=f(x) \text { and } G^{\prime}(x)=g(x)$$

1.4.2. 按部分集成。我们在微积分中遇到的研究最多的公式之一是两个导数的乘积:
$$\frac{d}{d x} F(x) G(x)=F^{\prime}(x) G(x)+F(x) G^{\prime}(x)$$

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

$$F^{\prime}(x)=f(x) \text { and } G^{\prime}(x)=g(x)$$

## 微积分作业代写calclulus代考|Preview

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

• $f$ 在每一点上都是连续的 $[a, b]$.
• $f$ 在每一点上都是连续的 $(a, b)$ 并且是有界的。
• $f$ 在每一点上都是连续的 $(a, b)$ 除了可能的一些点序列和 $f$ 是有界的。
• $f$ 在每一点上都是连续的 $(a, b)$ 最多有有限多个例外，并由另一个函数支配 $g$ 为此
$$0 \leq f(x) \leq g(x) \quad(a<x<b)$$
在哪里 $g$ 是连续的 $(a, b)$ (再次允许有限多个例外) 和积分 $\int_{a}^{b} g(x) d x$ 假定存在。