# 数学代写|微积分代写Calculus代写|MATH141

## 数学代写|微积分代写Calculus代写|The Fundamental Theorem of Calculus

By constructing appropriate covers, Archimedes was able to compute areas and integrals in certain situations. For example, he knew that $\int_0^1 x^2 d x=1 / 3$. On the other hand, Archimedes was also able to compute tangent lines to certain curves and surfaces. However, he apparently had no idea that these two processes were intimately related, through the fundamental theorem of calculus. It was the discovery of the fundamental theorem, in the seventeenth century, that turned the computation of areas from a mystery to a simple and straightforward reality.

In this section, all functions will be continuous. Since we will use $f^{+}$and $f^{-}$repeatedly, it is important to note that $(\S 2.3)$ a function is continuous iff both its positive and negative parts are continuous.

Let $f$ be continuous on $(a, b)$, and let $[c, d]$ be a compact subinterval. Since (§2.3) continuous functions map compact intervals to compact intervals, $f$ is bounded on $[c, d]$, hence, integrable over $(c, d)$.
Let $f$ be continuous on $(a, b)$, fix $a<c<b$, and set
$$F_c(x)=\left{\begin{array}{lc} \int_c^x f(t) d t, & c \leq x<b, \ -\int_x^c f(t) d t, & a<x \leq c . \end{array}\right.$$
By the previous paragraph, $F_c(x)$ is finite for all $a<x<b$. From the previous section, we know that $F_c$ is continuous. Here, we show that $F_c$ is differentiable and $F_c^{\prime}(x)=f(x)$ on $(a, b)$ (Figure 4.27). We will need the modulus of continuity $\mu_x(\S 2.3)$ of $f$ at $x$. To begin, by additivity, $F_c(y)-$ $F_c(x)=F_x(y)-F_x(x)$ for any two points $x, y$ in $(a, b)$, whether they are to the right or the left of $c$.

Then, for $a<x<t<y<b,|f(t)-f(x)| \leq \mu_x(y-x)$. Thus, $f(t) \leq$ $f(x)+\mu_x(y-x)$. Hence,
\begin{aligned} \frac{F_c(y)-F_c(x)}{y-x} &=\frac{F_x(y)-F_x(x)}{y-x} \ &=\frac{1}{y-x} \int_x^y f(t) d t \ & \leq \frac{1}{y \quad x} \int_x^y\left[f(x)+\mu_x(y-x)\right] d t=f(x)+\mu_x(y-x) . \end{aligned}
Similnrly, since $a<x<t<y<b$ implins $f(x) \quad \mu_x(y \quad x) \leq f(t)$
$$\frac{F_c(y)-F_c(x)}{y-x} \geq f(x)-\mu_x(y-x) .$$

## 数学代写|微积分代写Calculus代写|The Method of Exhaustion

In this section, we compute the area of the unit disk $D$ via the Method of Exhaustion.

For $n \geq 3$, let $P_k=(\cos (2 \pi k / n), \sin (2 \pi k / n)), 0 \leq k \leq n$. Then, the points $P_k$ are evenly spaced about the unit circle $\left{(x, y): x^2+y^2=1\right}$, and $P_n=P_0$. Let $D_n \subset D$ be the interior of the inscribed regular $n$-sided polygon obtained by joining the points $P_0, P_1, \ldots, P_n$ (we do not include the edges of $D_n$ in the definition of $D_n$ ). Then (Exercise 4.2.13),

$$\text { area }\left(D_n\right)=\frac{n}{2} \sin (2 \pi / n)=\pi \cdot \frac{\sin (2 \pi / n)}{2 \pi / n} \text {. }$$
Since $\lim {x \rightarrow 0} \frac{\sin x}{x}=\sin ^{\prime} 0=\cos 0=1$, we obtain $$\lim {n \zeta \infty} \operatorname{area}\left(D_n\right)=\pi .$$
Since
$$D_4 \subset D_8 \subset D_{16} \subset \ldots, \quad \text { and } \quad D=\bigcup_{n=2}^{\infty} D_{2^n},$$
it is reasonable to make the guess that
$$\operatorname{area}(D)=\lim {n>\infty} \operatorname{area}\left(D{2^n}\right),$$
and, hence, conclude that area $(D)=\pi$. The reasoning that leads from (4.5.2) to (4.5.3) is generally correct. The result is called the Method of Exhaustion. Although area $(D)$ was computed in the previous section using the fundamental theorem, in Chapter 5 we will need the Method to compute other areas.

We say that a sequence of sets $\left(A_n\right)$ is increasing (Figure 4.29) if $A_1 \subset$ $A_2 \subset A_3 \subset \ldots$

# 微积分代考

## 数学代写|微积分代写Calculus代写|微积分基本定理

$$F_c(x)=\left{\begin{array}{lc} \int_c^x f(t) d t, & c \leq x<b, \ -\int_x^c f(t) d t, & a<x \leq c . \end{array}\right.$$

Then, for $a<x<t<y<b,|f(t)-f(x)| \leq \mu_x(y-x)$。因此，f(t) \leq$$f(x)+\mu_x(y-x)。因此，$$ \begin{aligned} \frac{F_c(y)-F_c(x)}{y-x} &=\frac{F_x(y)-F_x(x)}{y-x} \ &=\frac{1}{y-x} \int_x^y f(t) d t \ & \leq \frac{1}{y \quad x} \int_x^y\left[f(x)+\mu_x(y-x)\right] d t=f(x)+\mu_x(y-x) . \end{aligned} $$同样，因为a<x<t<y<b隐含f(x) \quad \mu_x(y \quad x) \leq f(t)$$ \frac{F_c(y)-F_c(x)}{y-x} \geq f(x)-\mu_x(y-x) . $$## 数学代写|微积分代写Calculus代写|The Method of depletion 在本节中，我们通过用尽法计算单位硬盘的面积D 对于n \geq 3，让P_k=(\cos (2 \pi k / n), \sin (2 \pi k / n)), 0 \leq k \leq n。然后，点P_k围绕单位圆\left{(x, y): x^2+y^2=1\right}和P_n=P_0均匀间隔。设D_n \subset D是通过连接P_0, P_1, \ldots, P_n点(我们在D_n的定义中不包括D_n的边)获得的n边正多边形的内部。然后(练习4.2.13)，$$ \text { area }\left(D_n\right)=\frac{n}{2} \sin (2 \pi / n)=\pi \cdot \frac{\sin (2 \pi / n)}{2 \pi / n} \text {. } $$由于\lim {x \rightarrow 0} \frac{\sin x}{x}=\sin ^{\prime} 0=\cos 0=1，我们得到$$ \lim {n \zeta \infty} \operatorname{area}\left(D_n\right)=\pi . $$由于$$ D_4 \subset D_8 \subset D_{16} \subset \ldots, \quad \text { and } \quad D=\bigcup_{n=2}^{\infty} D_{2^n}, $$，可以合理地作出猜测，$$ \operatorname{area}(D)=\lim {n>\infty} \operatorname{area}\left(D{2^n}\right), $$，因此，得出结论，区域(D)=\pi。从(4.5.2)到(4.5.3)的推理一般是正确的。其结果被称为用尽法。虽然在上一节中使用基本定理计算了区域(D)，但在第5章中我们将需要该方法来计算其他区域 如果A_1 \subset$$A_2 \subset A_3 \subset \ldots

，我们说集合的序列$\left(A_n\right)$正在增加(图4.29)

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