### 物理代写|理论力学作业代写Theoretical Mechanics代考|Elements of Integral Calculus

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## 物理代写|理论力学作业代写Theoretical Mechanics代考|Notions

The technique of ‘differentiation’, which we discussed in the previous section, follows the scope of work:
given: $\quad y=f(x)$
finding: $\quad f^{\prime}(x)=\frac{d f}{d x}:$ ‘derivation’,
The reverse program, namely
given: $\quad f^{\prime}(x)=\frac{d f}{d x}$
finding: $\quad y=f(x)$
leads to the technique of ‘integration’. Consider for example
$$f^{\prime}(x)=c=\text { const }$$
then we remember according to (1.77) that
$$y=f(x)=c \cdot x$$
fulfills the condition $f^{\prime}(x)=c$.
Definition $F(x)$ is the ‘antiderivative (primitive function)’ of $f(x)$, if it holds:
$$F^{\prime}(x)=f(x) \quad \forall x$$
In this connection the above example means:
$$f(x) \equiv c \quad \curvearrowright \quad F(x)=c \cdot x+d .$$

Because of the constant $d$ the result comes out as a full family of curves. Fixing $d$ needs the introduction of ‘boundary conditions’. We accept that:
‘Integration’ : Searching for the antiderivative (primitive function)
To generate a graphic image, the integral can be interpreted as the area under the curve $y=f(x)$. If the curve $y=f(x)$ is given then we ask ourselves how we can determine the area $F$ in Fig. $1.18$ under the curve between the limits $x=$ $a$ and $x=b$. This can easily be done for the special case that $f(x)$ represents a straight line. However, how can we calculate the area under an arbitrary (continuous) function $f(x)$ ?

In a first step, we approach the calculation of the area by decomposing the interval $[a, b]$ in $n$ equal sub-intervals $\Delta x_{n}$,
$$\Delta x_{n}=\frac{b-a}{n} \quad n \in \mathbb{N} / 0$$
where $x_{i}$ is the center of the $i$-th partial interval:
$$x_{i}=a+\left(i-\frac{1}{2}\right) \Delta x_{n} ; i=1,2 \ldots, n$$
Then $f\left(x_{i}\right) \Delta x_{n}$ is the area of the $i$-th pillar in Fig. 1.19. Hence it holds approximately for the area $F$ :
$$F \approx \sum_{i=1}^{n} f\left(x_{i}\right) \Delta x_{n}$$

## 物理代写|理论力学作业代写Theoretical Mechanics代考|First Rules of Integration

Some important rules follow directly from the definition of the integral:

• Identical bounds of integration:
$$\int_{a}^{a} f(x) d x=0$$
• The ‘area’ in the sense of an integral has a sign because of $f(x)<^{\geq}$! For the example in Fig. $1.20$ one recognizes: \begin{aligned} &F_{1}=\int_{a}^{x_{1}} f(x) d x>0 \ &F_{2}=\int_{x_{1}}^{x_{2}} f(x) d x<0 \\ &F_{3}=\int_{x_{2}}^{x_{3}} f(x) d x>0 \end{aligned}Constant factor $c \in \mathbb{R}$ :
• $$• \int_{a}^{b} c \cdot f(x) d x=\lim {n \rightarrow \infty} \sum{i=1}^{n} c \cdot f\left(x_{i}\right) \Delta x_{n}=c \cdot \lim {n \rightarrow \infty} \sum{i=1}^{n} f\left(x_{i}\right) \Delta x_{n} •$$
• Hence it holds:
• $$• \int_{a}^{b} c \cdot f(x) d x=c \cdot \int_{a}^{b} f(x) d x •$$
• Sum:
• Assume
• $$• f(x) \equiv g(x)+h(x) •$$
• then it follows from the definition of the Riemann integral:
• • \begin{aligned} • \int_{a}^{b} f(x) d x &=\lim {n \rightarrow \infty} \sum{i=1}^{n}\left(g\left(x_{i}\right)+h\left(x_{i}\right)\right) \Delta x_{n} \ • &=\lim {n \rightarrow \infty} \sum{i=1}^{n} g\left(x_{i}\right) \Delta x_{n}+\lim {n \rightarrow \infty} \sum{i=1}^{n} h\left(x_{i}\right) \Delta x_{n} • \end{aligned} •
• That means:
• $$• \int_{a}^{b} f(x) d x=\int_{a}^{b} g(x) d x+\int_{a}^{b} h(x) d x •$$
• The last two rules of integration (1.112) and (1.113) demonstrate the linearity of the integral.
• Partitioning the interval of integration:
• $$• \Delta x_{n}=\frac{b-a}{n}=\Delta x_{n}^{(1)}+\Delta x_{n}^{(2)}=\frac{x_{0}-a}{n}+\frac{b-x_{0}}{n} •$$
• Therewith we can write:
• $$• \int_{a}^{b} f(x) d x=\lim {n \rightarrow \infty} \sum{i=1}^{n} f\left(x_{i}^{(1)}\right) \Delta x_{n}^{(1)}+\lim {n \rightarrow \infty} \sum{i=1}^{n} f\left(x_{i}^{(2)}\right) \Delta x_{n}^{(2)} •$$

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Fundamental Theorem of Calculus

We consider the definite integral over a continuous function $f(t)$, but now with variable upper limit:
$$F(x)=\int_{a}^{x} f(t) d t$$
‘area function’

The area under the curve $f(t)$ in this case is not constant but a function of $x$ (Fig. 1.22). If the upper bound of integration is shifted by $\Delta x$ the area will change by:
$$\Delta F=F(x+\Delta x)-F(x)=\int_{a}^{x+\Delta x} f(t) d t-\int_{a}^{x} f(t) d t=\int_{x}^{x+\Delta x} f(t) d t$$
In the last step we have used the rule (1.114). Without explicit proof we accept the important
‘mean value theorem of integral calculus’
This theorem implies:
$$\exists \hat{x} \in[x, x+\Delta x] \text { with } \Delta F=\Delta x \cdot f(\hat{x}) \text {. }$$
Although not exactly proven the theorem appears rather plausible according to Fig. 1.23. So we can further conclude:
$$F^{\prime}(x)=\lim {\Delta x \rightarrow 0} \frac{\Delta F}{\Delta x}=\lim {\Delta x \rightarrow 0} f(\hat{x})=f(x)$$
Thus after $(1.105)$, the area function is the antiderivative of $f(x)$ ! Furthermore, the equivalence of the definitions $(1.105)$ and $(1.110)$ for the antiderivative, which

remained unsettled in Sect. 1.2.1, is now settled.
‘fundamental theorem of calculus’
$$\frac{d}{d x} F(x) \equiv \frac{d}{d x} \int_{a}^{x} f(t) d t=f(x) .$$
The successive performing of integration and differentiation obviously cancel each other!
$$\text { integration } \cong \text { inversion of differentiation }$$
The influence of the lower limit of integration in (1.118) still appears unsettled (Fig. 1.24). To clarify this we therefore investigate:
$$\tilde{F}(x)=\int_{a^{\prime}}^{x} f(t) d t=\underbrace{\int_{a^{\prime}}^{a} f(t) d t}{=A, \text { independent of } x}+\underbrace{\int{a}^{x} f(t) d t}_{F(x)} .$$
Therewith it follows that both $F(x)$ and $\tilde{F}(x)$ are antiderivatives of $f(x)$ :
$$\tilde{F}(x)=F(x)+A \curvearrowright \frac{d}{d x} \tilde{F}(x)=\frac{d}{d x} F(x)=f(x) .$$
The lower limit of integration is therefore in a certain sense dummy, the antiderivative is uniquely fixed except for an additive constant:
$$F(x) \Longleftrightarrow \tilde{F}(x)=F(x)+A .$$
Therefore one introduces the
$$\text { ‘indefinite integral’: } \quad F(x)=\int f(x) d x \text {. }$$
defining therewith the set of all antiderivatives of $f(x)$ !

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Notions

F′(X)=C= 常量

F′(X)=F(X)∀X

F(X)≡C↷F(X)=C⋅X+d.

“积分”：搜索反导数（原始函数）

ΔXn=b−一种nn∈ñ/0

X一世=一种+(一世−12)ΔXn;一世=1,2…,n

F≈∑一世=1nF(X一世)ΔXn

## 物理代写|理论力学作业代写Theoretical Mechanics代考|First Rules of Integration

• 相同的积分界限：
∫一种一种F(X)dX=0
• 积分意义上的“面积”有一个符号，因为F(X)<≥！对于图中的示例。1.20一个人承认：F1=∫一种X1F(X)dX>0 F2=∫X1X2F(X)dX<0F3=∫X2X3F(X)dX>0常数因子C∈R:
• $$• \int_{a}^{b} c \cdot f(x) dx=\lim {n \rightarrow \infty} \sum{i=1}^{n} c \cdot f\left(x_{i}\右) \Delta x_{n}=c \cdot \lim {n \rightarrow \infty} \sum{i=1}^{n} f\left(x_{i}\right) \Delta x_{n} •$$
• 因此它成立：
• $$• \int_{a}^{b} c \cdot f(x) dx=c \cdot \int_{a}^{b} f(x) dx •$$
• 和：
• 认为
• $$• f(x) \equiv g(x)+h(x) •$$
• 则由黎曼积分的定义得出：
• $$• \开始{对齐} • \int_{a}^{b} f(x) dx &=\lim {n \rightarrow \infty} \sum{i=1}^{n}\left(g\left(x_{i}\right) +h\left(x_{i}\right)\right) \Delta x_{n} \ • &=\lim {n \rightarrow \infty} \sum{i=1}^{n} g\left(x_{i}\right) \Delta x_{n}+\lim {n \rightarrow \infty} \ sum{i=1}^{n} h\left(x_{i}\right) \Delta x_{n} • \end{对齐} •$$
• 这意味着：
• $$• \int_{a}^{b} f(x) dx=\int_{a}^{b} g(x) d x+\int_{a}^{b} h(x) dx •$$
• 最后两个积分规则 (1.112) 和 (1.113) 证明了积分的线性。
• 划分积分区间：
• $$• \Delta x_{n}=\frac{ba}{n}=\Delta x_{n}^{(1)}+\Delta x_{n}^{(2)}=\frac{x_{0}- a}{n}+\frac{b-x_{0}}{n} •$$
• 我们可以这样写：
• $$• \int_{a}^{b} f(x) dx=\lim {n \rightarrow \infty} \sum{i=1}^{n} f\left(x_{i}^{(1)}\右) \Delta x_{n}^{(1)}+\lim {n \rightarrow \infty} \sum{i=1}^{n} f\left(x_{i}^{(2)}\右）\Delta x_{n}^{(2)} •$$

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Fundamental Theorem of Calculus

F(X)=∫一种XF(吨)d吨
‘面积函数’

ΔF=F(X+ΔX)−F(X)=∫一种X+ΔXF(吨)d吨−∫一种XF(吨)d吨=∫XX+ΔXF(吨)d吨

∃X^∈[X,X+ΔX] 和 ΔF=ΔX⋅F(X^).

F′(X)=林ΔX→0ΔFΔX=林ΔX→0F(X^)=F(X)

‘微积分基本定理’

ddXF(X)≡ddX∫一种XF(吨)d吨=F(X).

一体化 ≅ 微分反转
The influence of the lower limit of integration in (1.118) still appears unsettled (Fig. 1.24). To clarify this we therefore investigate:
$$\tilde{F}(x)=\int_{a^{\prime}}^{x} f(t) d t=\underbrace{\int_{a^{\prime}}^{a} f(t) d t}{=A, \text { independent of } x}+\underbrace{\int{a}^{x} f(t) d t}_{F(x)} . TherewithitfollowsthatbothF(x)andF~(x)areantiderivativesoff(x): \tilde{F}(x)=F(x)+A \curvearrowright \frac{d}{d x} \tilde{F}(x)=\frac{d}{d x} F(x)=f(x) . Thelowerlimitofintegrationisthereforeinacertainsensedummy,theantiderivativeisuniquelyfixedexceptforanadditiveconstant: F(x) \Longleftrightarrow \tilde{F}(x)=F(x)+A . Thereforeoneintroducesthe \text { ‘indefinite integral’: } \quad F(x)=\int f(x) d x \text {. }$$
defining therewith the set of all antiderivatives of f(x) !

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