Posted on Categories:Real analysis, 实分析, 数学代写

# 数学代写|实分析代写Real Analysis代考|Integration by substitution

avatest™

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

## 数学代写|实分析代写Real Analysis代考|Integration by substitution

Let (i) $f(a, b]-\mathbb{R}$ be integrable on $[a, b]$, $a \phi(\beta)=b, a n d$
(iii) $f \circ \phi$ and $\phi^{\prime}$ are integrable on $[\alpha, \beta]$ and $\phi^{\prime}(t) \neq 0$ for all $t, \theta$ $[\alpha, \beta]$
Then $\int_a^b f(x) d x=\int_\alpha^\beta f(\phi(t)) \phi^{\prime}(t) d t$
Since $\phi^{\prime}(t) \neq 0$ on $[\alpha, \beta]$, it follows from Darboux’s theorem that either $\phi^{\prime}(t)>0$ for all $t \in[\alpha, \beta]$ or $\phi^{\prime}(t)<0$ for all $t \in[\alpha, \beta]$, i.e., either $\phi$ is strictly increasing on $[\alpha, \beta]$ or $\phi$ is strictly decreasing on $[\alpha, \beta]$.
Accordingly, the theorem can be stated in two parts.
First part.
Let (i) $f:[a, b] \rightarrow \mathbb{R}$ be integrable on $[a, b]$,
(ii) $\phi:[\alpha, \beta] \rightarrow \mathbb{R}$ be differentiable and strictly increasing on $[\alpha, \beta]$ such that $\phi(\alpha)=a, \phi(\beta)=b$, and
(iii) $f \circ \phi$ and $\phi^{\prime}$ are integrable on $[\alpha, \beta]$.
Then $\int_a^b f(x) d x=\int_\alpha^\beta f(\phi(t)) \phi^{\prime}(t) d t$.
Proof. Since $\phi$ is differentiable on $[\alpha, \beta], \phi$ is continuous on $[\alpha, \beta]$.
Since $\phi$ is continuous and strictly increasing on $[\alpha, \beta]$ and $\phi(\alpha)=$ $a, \phi(\beta)=b, \phi^{-1}$ is continuous and strictly increasing on $[a, b]$.

Let $P=\left(x_0, x_1, \ldots, x_n\right)$ be any partition of $[a, b]$ and $Q=$ $\left{y_0, y_1, \ldots, y_n\right}$ where $y_i=\phi^{-1}\left(x_i\right)$ be the corresponding partition of $[\alpha, \beta]$.

By Lagrange’s Mean value theorem for the function $\phi$ on $\left[y_{r-1}, y_r\right]$, $\phi\left(y_r\right)-\phi\left(y_{r-1}\right)=\left(y_r-y_{r-1}\right) \phi^{\prime}\left(r_r\right)$ for some $r_r \in\left(y_{r-1}, y_r\right)$.
That is, $x_r-x_{r-1}=\left(y_r-y_{r-1}\right) \phi^{\prime}\left(r_r\right), r=1,2, \ldots, n \ldots$
Let $\phi\left(\eta_r\right)=\xi_r, r=1,2, \ldots, n$.
Now $S(P, f, \xi)=f\left(\xi_1\right)\left(x_1-x_0\right)+f\left(\xi_2\right)\left(x_2-x_1\right)+\cdots+f\left(\xi_n\right)\left(x_n-x_{n-1}\right)$ $=f\left(\phi\left(\eta_1\right)\right) \phi^{\prime}\left(\eta_1\right)\left(y_1-y_0\right)+\cdots+f\left(\phi\left(\eta_n\right)\right) \phi^{\prime}\left(\eta_n\right)\left(y_n-y_{n-1}\right)$ $=S\left(Q,(f \circ \phi) \cdot \phi^{\prime}, \eta\right)$.
Since $f$ is integrable on $[a, b], \lim _{|P| \rightarrow 0} S(P, f, \xi)=\int_a^b f$.

## 数学代写|实分析代写Real Analysis代考|Integration by parts

Theorem 11.11.1. Let $f:[a, b] \rightarrow \mathbb{R}, g:[a, b] \rightarrow \mathbb{R}$ be both differentiable on $[a, b]$ and $f^{\prime}, g^{\prime}$ are both integrable on $[a, b]$. Then
$$\int_a^b f(x) g^{\prime}(x) d x=f(b) g(b)-f(a) g(a)-\int_a^b f^{\prime}(x) g(x) d x .$$
Proof. Since $f$ and $g$ are both differentiable on $[a, b], f g$ is differentiable on $[a, b]$.

Since $f$ and $g$ are differentiable on $[a, b], f$ and $g$ are continuous on $[a, b]$ and therefore they are both integrable on $[a, b]$.

Therefore $f g^{\prime}+f^{\prime} g$ is integrable on $[a, b]$, i.e., $(f g)^{\prime}$ is integrable on ${a, b]$.

So by the fundamental theorem, $\int_a^b\left(f^{\prime} g\right)^{\prime}=[f g]_a^b=f(b) g(b)-$ $f(a) g(a)$. Also $\int_a^b(f g)^{\prime}=\int_a^b\left(f g^{\prime}+f^{\prime} g\right)=\int_a^b f g^{\prime}+\int_a^b f^{\prime} g$.
Therefore $\int_a^b f(x) g^{\prime}(x) d x+\int_a^b f^{\prime}(x) g(x) d x=f(b) g(b)-f(a) g(a)$ or, $\int_a^b f(x) g^{\prime}(x) d x=f(b) g(b)-f(a) g(a)-\int_a^b f^{\prime}(x) g(x) d x$.

Theorem 11.12.1. (First Mean value theorem)
If (i) $f:[a, b] \rightarrow \mathbb{R}$ and $g:[a, b] \rightarrow \mathbb{R}$ be both integrable on $[a, b]$, and
(ii) $g(x)$ has the same sign for all $x \in[a, b]$,
then there is a number $\mu$ such that
$\int_a^b f(x) g(x) d x=\mu \int_a^b g(x) d x$ where $m \leq \mu \leq M$ and
$$m=\inf {x \in[a, b]} f(x), M=\sup {x \in[a, b]} g(x) \text {. }$$
If further, $f$ is continuous on $[a, b]$ then there exists a point $\xi$ in $[a, b]$ such that $\int_a^b f(x) g(x) d x=f(\xi) \int_a^b g(x) d x$.
Proof. Case 1. Let $g(x)>0, x \in[a, b]$.
Since $m=\inf {x \in[a, b]} f(x)$ and $M=\sup {x \in[a, b]} f(x), m \leq f(x) \leq M$ for all $x \in[a, b]$. Therefore $m g(x) \leq f(x) g(x) \leq M g(x)$ for all $x \in[a, b]$.

Since $f$ and $g$ are both integrable on $[a, b], m g, f g$ and $M g$ are integrable on $[a, b]$, and
\begin{aligned} & \int_n^b m g(x) d x \leq \int_a^b f(x) g(x) d x \leq \int_a^b M g(x) d x \ & \text { or, } m \int_a^b g(x) d x \leq \int_a^b f(x) g(x) d x \leq M \int_a^b g(x) d x . \end{aligned}
Therefore $\int_a^b f(x) g(x) d x=\mu \int_a^b g(x) d x$, where $m \leq \mu \leq M$.

## 数学代写|实分析代写Real Analysis代考|Integration by substitution

(iii) $f \circ \phi$ 和 $\phi^{\prime}$ 是可积的 $[\alpha, \beta]$ 和 $\phi^{\prime}(t) \neq 0$ 对所有人 $t, \theta$ $[\alpha, \beta]$

(ii) $\phi:[\alpha, \beta] \rightarrow \mathbb{R}$ 是可微且严格递增的 $[\alpha, \beta]$ 这样 $\phi(\alpha)=a, \phi(\beta)=b$，和
(iii) $f \circ \phi$ 和 $\phi^{\prime}$ 是可积的 $[\alpha, \beta]$．

(ii) $g(x)$对所有$x \in[a, b]$具有相同的符号，

$\int_a^b f(x) g(x) d x=\mu \int_a^b g(x) d x$其中$m \leq \mu \leq M$和
$$m=\inf {x \in[a, b]} f(x), M=\sup {x \in[a, b]} g(x) \text {. }$$

\begin{aligned} & \int_n^b m g(x) d x \leq \int_a^b f(x) g(x) d x \leq \int_a^b M g(x) d x \ & \text { or, } m \int_a^b g(x) d x \leq \int_a^b f(x) g(x) d x \leq M \int_a^b g(x) d x . \end{aligned}

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。