Unit B5: Further Applications: Work, Center of Mass, Probability第 B5 单元：积分的进一步应用：功、质心、概率

A force of $40\,\text{N}$ holds a spring stretched $0.2\,\text{m}$ beyond its natural length. Find the work done in stretching it from $0.2\,\text{m}$ to $0.5\,\text{m}$ beyond natural length.

First find $k$ from Hooke's law: $40 = k(0.2)$, so $k = 200\,\text{N/m}$. Then

$$ W = \int_{0.2}^{0.5} 200x\,dx = 100x^2\Big|_{0.2}^{0.5} = 100(0.25 - 0.04) = 21\,\text{J}. $$

一个力 $40\,\text{N}$ 将弹簧从自然长度拉伸 $0.2\,\text{m}$。求将其从 $0.2\,\text{m}$ 拉伸至 $0.5\,\text{m}$ 所做的功。

由胡克定律求 $k$：$40 = k(0.2)$，故 $k = 200\,\text{N/m}$。则

$$ W = \int_{0.2}^{0.5} 200x\,dx = 100x^2\Big|_{0.2}^{0.5} = 100(0.25 - 0.04) = 21\,\text{J}. $$

A cylindrical tank of radius $2\,\text{m}$ and height $5\,\text{m}$ is full of water (weight density $9800\,\text{N/m}^3$). Find the work to pump all the water out over the top rim. Let $y$ measure height above the base.

A slab at height $y$ has volume $\pi(2)^2\,dy = 4\pi\,dy$ and must rise a distance $5 - y$ to clear the rim:

$$ W = \int_0^5 9800\,(4\pi)(5-y)\,dy = 39200\pi\int_0^5 (5-y)\,dy = 39200\pi\Big[5y - \tfrac{y^2}{2}\Big]_0^5. $$ $$ W = 39200\pi\big(25 - 12.5\big) = 490000\pi \approx 1.54\times 10^{6}\,\text{J}. $$

一个半径 $2\,\text{m}$、高 $5\,\text{m}$ 的圆柱形水箱装满水（重度 $9800\,\text{N/m}^3$）。求将所有水从顶部抽出所做的功。设 $y$ 为距底部的高度。

高度 $y$ 处的薄片体积为 $\pi(2)^2\,dy = 4\pi\,dy$，需提升距离 $5 - y$ 才能越过顶部：

$$ W = \int_0^5 9800\,(4\pi)(5-y)\,dy = 39200\pi\int_0^5 (5-y)\,dy = 39200\pi\Big[5y - \tfrac{y^2}{2}\Big]_0^5. $$ $$ W = 39200\pi\big(25 - 12.5\big) = 490000\pi \approx 1.54\times 10^{6}\,\text{J}. $$

An inverted right-circular cone (point down) has height $10\,\text{m}$ and top radius $4\,\text{m}$. It is filled to a depth of $8\,\text{m}$ with water ($w = 9800\,\text{N/m}^3$). Find the work to pump all the water to the top rim of the cone. Let $y$ measure height above the vertex.

The cone widens linearly from radius $0$ at $y=0$ to radius $4$ at $y=10$, so the radius at height $y$ is $r(y) = \tfrac{4}{10}y = \tfrac{2}{5}y$. A slab at height $y$ has area $A(y) = \pi r^2 = \pi\big(\tfrac{2}{5}y\big)^2 = \tfrac{4\pi}{25}y^2$. The water occupies $0 \le y \le 8$, and each slab must rise to the rim at $y=10$, a distance $10 - y$:

$$ W = \int_0^8 9800\cdot \tfrac{4\pi}{25}y^2 (10 - y)\,dy = \frac{39200\pi}{25}\int_0^8 \big(10y^2 - y^3\big)\,dy. $$ $$ \int_0^8 \big(10y^2 - y^3\big)\,dy = \Big[\tfrac{10}{3}y^3 - \tfrac14 y^4\Big]_0^8 = \tfrac{10}{3}(512) - \tfrac14(4096) = \tfrac{5120}{3} - 1024 = \tfrac{2048}{3}. $$ $$ W = \frac{39200\pi}{25}\cdot \frac{2048}{3} = \frac{80281600\pi}{75} \approx 3.36\times 10^{6}\,\text{J}. $$

The lesson: when the tank is not a cylinder, $A(y)$ is a genuine function of $y$, and the lift distance is measured to wherever the water exits, not to the top of the fluid.

一个倒置的直圆锥（顶点朝下），高 $10\,\text{m}$，顶部半径 $4\,\text{m}$，装水至深度 $8\,\text{m}$（水的重度 $w = 9800\,\text{N/m}^3$）。求将全部水抽至顶部边缘所做的功。设 $y$ 为距顶点的高度。

锥体半径从 $y=0$ 时的 $0$ 线性增大到 $y=10$ 时的 $4$，故高度 $y$ 处的半径为 $r(y) = \tfrac{2}{5}y$。薄片面积 $A(y) = \tfrac{4\pi}{25}y^2$。水占据 $0 \le y \le 8$，每层薄片须提升至 $y=10$ 处，距离为 $10 - y$：

$$ W = \int_0^8 9800\cdot \tfrac{4\pi}{25}y^2 (10 - y)\,dy = \frac{39200\pi}{25}\int_0^8 \big(10y^2 - y^3\big)\,dy. $$ $$ \int_0^8 \big(10y^2 - y^3\big)\,dy = \Big[\tfrac{10}{3}y^3 - \tfrac14 y^4\Big]_0^8 = \tfrac{5120}{3} - 1024 = \tfrac{2048}{3}. $$ $$ W = \frac{39200\pi}{25}\cdot \frac{2048}{3} = \frac{80281600\pi}{75} \approx 3.36\times 10^{6}\,\text{J}. $$

关键：当容器不是圆柱体时，$A(y)$ 是 $y$ 的真实函数，提升距离应量至液体出口处，而非液面处。

A uniform cable $20\,\text{m}$ long with linear density $3\,\text{kg/m}$ hangs from the top of a building. Find the work to wind the entire cable up to the top. Take $g = 9.8\,\text{m/s}^2$.

Let $x$ be the length of cable already wound up, so $0 \le x \le 20$. When $x$ meters have been raised, the still-hanging portion has length $20 - x$, mass $3(20 - x)$, and weight $3(20 - x)g$. Raising it a further $dx$ does work $3(20-x)g\,dx$. Equivalently, slice the cable: a piece at depth $x$ below the top must be lifted distance $x$, contributing $g\cdot 3\,dx \cdot x$. Both viewpoints give

$$ W = \int_0^{20} 3g\,x\,dx = 3(9.8)\Big[\tfrac{x^2}{2}\Big]_0^{20} = 29.4\cdot 200 = 5880\,\text{J}. $$

If in addition a $50\,\text{kg}$ load hangs at the bottom and is lifted the full $20\,\text{m}$, it adds constant-force work $50(9.8)(20) = 9800\,\text{J}$, for a total of $15680\,\text{J}$. The cable contributes a variable force (it shortens as you wind), while the load contributes a constant force; the integral handles the first, simple multiplication the second.

一根均匀缆绳，长 $20\,\text{m}$，线密度 $3\,\text{kg/m}$，悬挂于楼顶。求将整根缆绳提升至顶部所做的功。取 $g = 9.8\,\text{m/s}^2$。

设 $x$ 为已卷起的缆绳长度，$0 \le x \le 20$。当提升了 $x$ 米时，剩余悬挂部分长 $20-x$，质量 $3(20-x)$，重力 $3(20-x)g$。再提升 $dx$ 做功 $3(20-x)g\,dx$。等价地，切片：顶部以下深度 $x$ 处的一小段需提升距离 $x$，贡献 $g\cdot 3\,dx\cdot x$。两种视角均给出：

$$ W = \int_0^{20} 3g\,x\,dx = 3(9.8)\Big[\tfrac{x^2}{2}\Big]_0^{20} = 29.4\cdot 200 = 5880\,\text{J}. $$

若底部还挂有 $50\,\text{kg}$ 的重物并被提升 $20\,\text{m}$，还需加上常力做功 $50(9.8)(20) = 9800\,\text{J}$，总共 $15680\,\text{J}$。缆绳提供变力（随卷起而缩短），重物提供常力；积分处理前者，乘法处理后者。

A rectangular dam wall is $10\,\text{m}$ wide and the water stands $6\,\text{m}$ deep against it. Find the total hydrostatic force ($w = 9800\,\text{N/m}^3$). Let $h$ be depth below the surface.

Each strip has constant width $L = 10$, so

$$ F = \int_0^6 9800\,h\,(10)\,dh = 98000\int_0^6 h\,dh = 98000\cdot \frac{36}{2} = 1.764\times 10^{6}\,\text{N}. $$

一矩形大坝墙宽 $10\,\text{m}$，水深 $6\,\text{m}$。求总静水压力（$w = 9800\,\text{N/m}^3$）。设 $h$ 为距液面的深度。

每条水平条宽度 $L = 10$（常数），故：

$$ F = \int_0^6 9800\,h\,(10)\,dh = 98000\int_0^6 h\,dh = 98000\cdot \frac{36}{2} = 1.764\times 10^{6}\,\text{N}. $$

Consider a thin column of fluid of cross-sectional area $A$ extending from the surface down to depth $h$. The fluid is in equilibrium, so the upward force on its base balances the weight of the column above it (plus atmospheric pressure, which we measure relative to).

The weight of the column is (density)(volume)(gravity) $= \rho (Ah) g$. Dividing by the base area $A$ gives the gauge pressure:

$$ P = \frac{\rho A h g}{A} = \rho g h = w h. $$

By Pascal's principle this pressure acts equally in all directions, so it presses with the same magnitude on a vertical wall at that depth.

考虑一截面积为 $A$、从液面延伸至深度 $h$ 的流体细柱。流体处于平衡，底面向上的力等于上方液柱的重力（相对大气压测量）。

液柱重力为（密度）乘以（体积）乘以（重力加速度）$= \rho (Ah) g$。除以底面积 $A$ 得到表压：

$$ P = \frac{\rho A h g}{A} = \rho g h = w h. $$

由帕斯卡定理（Pascal's principle），该压强向各方向均等作用，因此对同一深度的竖直墙面施加相同大小的压力。

A vertical gate is an isosceles triangle with horizontal top edge $4\,\text{m}$ wide and apex pointing down, total height $3\,\text{m}$. Its top edge lies at the water surface. Find the hydrostatic force ($w = 9800\,\text{N/m}^3$). Let $h$ measure depth below the surface, $0 \le h \le 3$.

The triangle narrows linearly from width $4$ at $h=0$ to width $0$ at $h=3$, so the strip width is $L(h) = 4\big(1 - \tfrac{h}{3}\big) = \tfrac{4}{3}(3 - h)$. The strip at depth $h$ feels pressure $wh$ and has area $L(h)\,dh$:

$$ F = \int_0^3 w\,h\cdot \tfrac{4}{3}(3 - h)\,dh = \frac{4w}{3}\int_0^3 \big(3h - h^2\big)\,dh. $$ $$ \int_0^3 \big(3h - h^2\big)\,dh = \Big[\tfrac{3h^2}{2} - \tfrac{h^3}{3}\Big]_0^3 = \tfrac{27}{2} - 9 = \tfrac{9}{2}. $$ $$ F = \frac{4(9800)}{3}\cdot \frac{9}{2} = 4(9800)\cdot \frac{3}{2} = 58800\,\text{N}. $$

The narrowing toward the bottom (where pressure is largest) keeps the force modest compared with a $4\,\text{m}$-wide rectangle of the same height.

一个竖直闸门为等腰三角形，顶部水平边宽 $4\,\text{m}$，顶点朝下，总高 $3\,\text{m}$，顶部边缘位于水面处。求静水压力（$w = 9800\,\text{N/m}^3$）。设 $h$ 为距水面深度，$0 \le h \le 3$。

三角形从 $h=0$ 时宽 $4$ 线性缩小到 $h=3$ 时宽 $0$，故条宽为 $L(h) = \tfrac{4}{3}(3 - h)$。深度 $h$ 处的水平条受压强 $wh$，面积 $L(h)\,dh$：

$$ F = \int_0^3 w\,h\cdot \tfrac{4}{3}(3 - h)\,dh = \frac{4w}{3}\int_0^3 \big(3h - h^2\big)\,dh. $$ $$ \int_0^3 \big(3h - h^2\big)\,dh = \Big[\tfrac{3h^2}{2} - \tfrac{h^3}{3}\Big]_0^3 = \tfrac{27}{2} - 9 = \tfrac{9}{2}. $$ $$ F = \frac{4(9800)}{3}\cdot \frac{9}{2} = 58800\,\text{N}. $$

底部越来越窄（而压强在底部最大），使得总力比同高度的 $4\,\text{m}$ 宽矩形更小。

A circular window of radius $1\,\text{m}$ is set into the vertical wall of an aquarium so that its center is $5\,\text{m}$ below the surface. Find the force on the window ($w = 9800\,\text{N/m}^3$). Place the origin at the center of the circle and let $y$ run from $-1$ (bottom) to $1$ (top), measured upward.

A horizontal strip at height $y$ lies at depth $h = 5 - y$ below the surface. The circle $y^2 + z^2 = 1$ has half-width $\sqrt{1 - y^2}$, so the strip width is $L = 2\sqrt{1 - y^2}$:

$$ F = \int_{-1}^{1} w(5 - y)\,2\sqrt{1 - y^2}\,dy = 2w\Big[5\!\int_{-1}^{1}\!\sqrt{1-y^2}\,dy \;-\; \int_{-1}^{1}\! y\sqrt{1-y^2}\,dy\Big]. $$

The first integral is the area of a unit semicircle's double, $\int_{-1}^{1}\sqrt{1-y^2}\,dy = \tfrac{\pi}{2}$. The second integrand is odd, so it vanishes by symmetry. Hence

$$ F = 2w\Big[5\cdot \tfrac{\pi}{2} - 0\Big] = 5\pi w = 5\pi(9800) \approx 1.54\times 10^{5}\,\text{N}. $$

Notice the answer equals $w\cdot(\text{depth of centroid})\cdot(\text{area}) = 9800\cdot 5\cdot \pi$, a shortcut we justify next.

一个半径 $1\,\text{m}$ 的圆形观察窗安装在水族馆竖直墙上，圆心位于液面以下 $5\,\text{m}$。求圆窗上的静水压力（$w = 9800\,\text{N/m}^3$）。以圆心为原点，$y$ 从 $-1$（底部）到 $1$（顶部），向上为正。

高度 $y$ 处的水平条位于深度 $h = 5 - y$。圆 $y^2 + z^2 = 1$ 的半宽为 $\sqrt{1-y^2}$，故条宽 $L = 2\sqrt{1-y^2}$：

$$ F = \int_{-1}^{1} w(5 - y)\,2\sqrt{1 - y^2}\,dy = 2w\Big[5\!\int_{-1}^{1}\!\sqrt{1-y^2}\,dy \;-\; \int_{-1}^{1}\! y\sqrt{1-y^2}\,dy\Big]. $$

第一个积分为单位半圆面积的两倍，$\int_{-1}^{1}\sqrt{1-y^2}\,dy = \tfrac{\pi}{2}$。第二个被积函数为奇函数，由对称性为零。因此：

$$ F = 2w\Big[5\cdot \tfrac{\pi}{2} - 0\Big] = 5\pi w = 5\pi(9800) \approx 1.54\times 10^{5}\,\text{N}. $$

注意结果等于 $w\cdot$（形心深度）$\cdot$（面积）$= 9800\cdot 5\cdot \pi$，这一捷径将在下文说明。

A rod occupies $0 \le x \le 2$ with density $\rho(x) = 1 + x$ (mass per unit length). Find its center of mass.

$$ m = \int_0^2 (1+x)\,dx = \Big[x + \tfrac{x^2}{2}\Big]_0^2 = 2 + 2 = 4. $$ $$ M = \int_0^2 x(1+x)\,dx = \int_0^2 (x + x^2)\,dx = \Big[\tfrac{x^2}{2} + \tfrac{x^3}{3}\Big]_0^2 = 2 + \tfrac{8}{3} = \tfrac{14}{3}. $$ $$ \bar{x} = \frac{M}{m} = \frac{14/3}{4} = \frac{7}{6}. $$

The balance point sits past the midpoint $x=1$, as expected since the rod is denser toward $x=2$.

细杆占据 $0 \le x \le 2$，密度 $\rho(x) = 1 + x$（单位长度质量）。求其质心。

$$ m = \int_0^2 (1+x)\,dx = \Big[x + \tfrac{x^2}{2}\Big]_0^2 = 4. $$ $$ M = \int_0^2 x(1+x)\,dx = \Big[\tfrac{x^2}{2} + \tfrac{x^3}{3}\Big]_0^2 = 2 + \tfrac{8}{3} = \tfrac{14}{3}. $$ $$ \bar{x} = \frac{M}{m} = \frac{14/3}{4} = \frac{7}{6}. $$

平衡点超过中点 $x=1$，符合预期，因为细杆向 $x=2$ 方向越来越密。

Three masses sit in the plane: $m_1 = 2$ at $(1,1)$, $m_2 = 3$ at $(2,-1)$, and $m_3 = 5$ at $(-1,2)$. Find the center of mass.

Total mass $m = 2 + 3 + 5 = 10$. The moment about the $y$-axis uses the $x$-coordinates, the moment about the $x$-axis uses the $y$-coordinates:

$$ M_y = \sum m_i x_i = 2(1) + 3(2) + 5(-1) = 2 + 6 - 5 = 3. $$ $$ M_x = \sum m_i y_i = 2(1) + 3(-1) + 5(2) = 2 - 3 + 10 = 9. $$ $$ (\bar{x}, \bar{y}) = \Big(\frac{M_y}{m}, \frac{M_x}{m}\Big) = \Big(\frac{3}{10}, \frac{9}{10}\Big). $$

The heavy mass at $(-1,2)$ pulls the balance up and to the left, which matches $\bar{y} = 0.9$ being high and $\bar{x} = 0.3$ being modest.

三个质点：$m_1 = 2$ 在 $(1,1)$，$m_2 = 3$ 在 $(2,-1)$，$m_3 = 5$ 在 $(-1,2)$。求质心。

总质量 $m = 10$。对 $y$ 轴的力矩用 $x$ 坐标，对 $x$ 轴的力矩用 $y$ 坐标：

$$ M_y = 2(1) + 3(2) + 5(-1) = 3. $$ $$ M_x = 2(1) + 3(-1) + 5(2) = 9. $$ $$ (\bar{x}, \bar{y}) = \Big(\frac{3}{10}, \frac{9}{10}\Big). $$

$(-1,2)$ 处的较重质量将平衡点向左上方拉动，与 $\bar{y} = 0.9$ 偏高、$\bar{x} = 0.3$ 居中的结果一致。

A rod occupies $0 \le x \le \pi$ with density $\rho(x) = \sin x$. Find its mass and center of mass.

$$ m = \int_0^\pi \sin x\,dx = \big[-\cos x\big]_0^\pi = 1 - (-1) = 2. $$

For the moment use integration by parts, $\int x\sin x\,dx = -x\cos x + \sin x$:

$$ M = \int_0^\pi x\sin x\,dx = \big[-x\cos x + \sin x\big]_0^\pi = \big(-\pi(-1) + 0\big) - (0 + 0) = \pi. $$ $$ \bar{x} = \frac{M}{m} = \frac{\pi}{2}. $$

Even though the density rises then falls, the rod is symmetric about $x = \tfrac{\pi}{2}$, so the balance point lands exactly at the center. This is the symmetry principle reading off an answer that the integral confirms.

细杆占据 $0 \le x \le \pi$，密度 $\rho(x) = \sin x$。求其质量和质心。

$$ m = \int_0^\pi \sin x\,dx = \big[-\cos x\big]_0^\pi = 2. $$

求力矩用分部积分法，$\int x\sin x\,dx = -x\cos x + \sin x$：

$$ M = \int_0^\pi x\sin x\,dx = \big[-x\cos x + \sin x\big]_0^\pi = \pi. $$ $$ \bar{x} = \frac{M}{m} = \frac{\pi}{2}. $$

虽然密度先升后降，但细杆关于 $x = \tfrac{\pi}{2}$ 对称，平衡点恰好在中心处。这是对称性原理直接给出答案，而积分验证了它。

The defining property of the center of mass $\bar{x}$ is that the system balances on a fulcrum placed there: the total signed moment about $\bar{x}$ is zero. Write that condition and solve.

$$ \sum_i m_i (x_i - \bar{x}) = 0 \;\Longrightarrow\; \sum_i m_i x_i = \bar{x}\sum_i m_i \;\Longrightarrow\; \bar{x} = \frac{\sum_i m_i x_i}{\sum_i m_i}. $$

So the formula is not a definition pulled from the air; it is forced by the requirement that moments cancel at the balance point. Replacing the sum by an integral, the same algebra gives $\bar{x} = \frac{1}{m}\int x\,dm$ for a continuous body, because $\int (x - \bar{x})\,dm = 0$ has the identical solution. The additivity of moments, $M_{\text{whole}} = \sum M_{\text{parts}}$, is just the additivity of integrals, and it is what lets you find the centroid of a complicated shape by splitting it into pieces.

质心 $\bar{x}$ 的定义性质是：系统恰好在该点的支点上平衡，即关于 $\bar{x}$ 的有符号总力矩为零。写出该条件并求解：

$$ \sum_i m_i (x_i - \bar{x}) = 0 \;\Longrightarrow\; \bar{x} = \frac{\sum_i m_i x_i}{\sum_i m_i}. $$

因此公式并非凭空定义，而是由平衡点处力矩相消的要求所决定。将求和换为积分，同样的代数运算给出连续体的 $\bar{x} = \frac{1}{m}\int x\,dm$。力矩的可叠加性 $M_{\text{whole}} = \sum M_{\text{parts}}$ 就是积分的可加性，正是这一性质使得我们可以将复杂形状分解成各部分来求形心（centroid）。

Find the centroid of the region under $y = x^2$ over $[0,1]$.

$$ A = \int_0^1 x^2\,dx = \tfrac13. $$ $$ \bar{x} = \frac{1}{A}\int_0^1 x\cdot x^2\,dx = 3\int_0^1 x^3\,dx = 3\cdot \tfrac14 = \tfrac34. $$ $$ \bar{y} = \frac{1}{A}\int_0^1 \tfrac12(x^2)^2\,dx = 3\cdot \tfrac12\int_0^1 x^4\,dx = \tfrac32\cdot \tfrac15 = \tfrac{3}{10}. $$

The centroid is $\big(\tfrac34, \tfrac{3}{10}\big)$, pulled right and low because the region is fat near $x=1$ and thin near $x=0$.

求 $[0,1]$ 上 $y = x^2$ 曲线下区域的形心。

$$ A = \int_0^1 x^2\,dx = \tfrac13. $$ $$ \bar{x} = 3\int_0^1 x^3\,dx = \tfrac34. \qquad \bar{y} = \tfrac32\int_0^1 x^4\,dx = \tfrac{3}{10}. $$

形心为 $\big(\tfrac34, \tfrac{3}{10}\big)$，偏右偏低，因为区域在 $x=1$ 附近较宽，在 $x=0$ 附近较窄。

Find the centroid of the region bounded above by $y = \sqrt{x}$ and below by $y = x$ on $[0,1]$.

On $[0,1]$, $\sqrt{x} \ge x$, so $f(x) = \sqrt{x}$ is the top and $g(x) = x$ is the bottom. The area is

$$ A = \int_0^1 \big(\sqrt{x} - x\big)\,dx = \Big[\tfrac{2}{3}x^{3/2} - \tfrac{x^2}{2}\Big]_0^1 = \tfrac{2}{3} - \tfrac12 = \tfrac{1}{6}. $$ $$ \bar{x} = \frac{1}{A}\int_0^1 x\big(\sqrt{x} - x\big)\,dx = 6\int_0^1 \big(x^{3/2} - x^2\big)\,dx = 6\Big[\tfrac{2}{5}x^{5/2} - \tfrac{x^3}{3}\Big]_0^1 = 6\big(\tfrac{2}{5} - \tfrac13\big) = 6\cdot \tfrac{1}{15} = \tfrac{2}{5}. $$ $$ \bar{y} = \frac{1}{A}\int_0^1 \tfrac12\big[(\sqrt{x})^2 - x^2\big]\,dx = 6\cdot \tfrac12\int_0^1 \big(x - x^2\big)\,dx = 3\Big[\tfrac{x^2}{2} - \tfrac{x^3}{3}\Big]_0^1 = 3\big(\tfrac12 - \tfrac13\big) = \tfrac12. $$

The centroid $\big(\tfrac25, \tfrac12\big)$ lies inside the sliver between the two curves, as it must.

求 $[0,1]$ 上 $y = \sqrt{x}$（上）与 $y = x$（下）所围区域的形心。

在 $[0,1]$ 上 $\sqrt{x} \ge x$。面积：

$$ A = \int_0^1 \big(\sqrt{x} - x\big)\,dx = \tfrac{2}{3} - \tfrac12 = \tfrac{1}{6}. $$ $$ \bar{x} = 6\int_0^1 \big(x^{3/2} - x^2\big)\,dx = 6\big(\tfrac{2}{5} - \tfrac13\big) = \tfrac{2}{5}. $$ $$ \bar{y} = 3\int_0^1 \big(x - x^2\big)\,dx = 3\big(\tfrac12 - \tfrac13\big) = \tfrac12. $$

形心 $\big(\tfrac25, \tfrac12\big)$ 位于两曲线之间的细长区域内，符合预期。

Find the centroid of the quarter-disk $x^2 + y^2 \le 1$ with $x \ge 0,\ y \ge 0$.

By the symmetry of swapping $x$ and $y$, the centroid satisfies $\bar{x} = \bar{y}$, so we only compute one. The region is bounded above by $y = \sqrt{1 - x^2}$ on $[0,1]$, and its area is $A = \tfrac{\pi}{4}$.

$$ \bar{x} = \frac{1}{A}\int_0^1 x\sqrt{1 - x^2}\,dx = \frac{4}{\pi}\Big[-\tfrac13(1 - x^2)^{3/2}\Big]_0^1 = \frac{4}{\pi}\cdot \tfrac13 = \frac{4}{3\pi}. $$

So the centroid is $\big(\tfrac{4}{3\pi}, \tfrac{4}{3\pi}\big) \approx (0.42, 0.42)$, comfortably inside the quarter-disk and biased toward the bulk of the region near the corner at the origin.

求四分之一圆盘 $x^2 + y^2 \le 1$（$x \ge 0,\ y \ge 0$）的形心。

由 $x$ 与 $y$ 互换的对称性，$\bar{x} = \bar{y}$，只需计算一个。上界为 $y = \sqrt{1-x^2}$，面积 $A = \tfrac{\pi}{4}$。

$$ \bar{x} = \frac{4}{\pi}\int_0^1 x\sqrt{1 - x^2}\,dx = \frac{4}{\pi}\cdot \tfrac13 = \frac{4}{3\pi}. $$

形心为 $\big(\tfrac{4}{3\pi}, \tfrac{4}{3\pi}\big) \approx (0.42, 0.42)$，位于四分之一圆盘内部，偏向原点附近质量集中的区域。

If a region is symmetric about a line $\ell$, then its centroid lies on $\ell$. The reason is a cancellation of moments: for every mass element at signed distance $+d$ from $\ell$ there is a mirror element at $-d$, so the total moment about $\ell$ is zero.

$$ M_{\ell} = \int (\text{signed distance})\,dm = 0 \implies \text{centroid on } \ell. $$

This lets you read off one coordinate of the centroid for free. A disk, a square, and an ellipse all have their centroid at their geometric center by two perpendicular symmetries.

若区域关于直线 $\ell$ 对称，则其形心（centroid）在 $\ell$ 上。原因是力矩相互抵消：对 $\ell$ 有符号距离 $+d$ 的每个质量微元，都有对应的 $-d$ 处的镜像微元，因此关于 $\ell$ 的总力矩为零。

$$ M_{\ell} = \int (\text{有符号距离})\,dm = 0 \implies \text{形心在 } \ell \text{ 上}. $$

这让我们可以免费读出形心的一个坐标。圆盘、正方形和椭圆都因两个垂直对称轴而使形心位于几何中心。

Find the average value of $f(x) = x^2$ on $[0,3]$, and a point $c$ where it is attained.

$$ f_{\text{avg}} = \frac{1}{3-0}\int_0^3 x^2\,dx = \frac{1}{3}\cdot \frac{27}{3} = 3. $$

Solve $c^2 = 3$ for $c$ in $[0,3]$: $c = \sqrt{3} \approx 1.73$, which lies in the interval as the theorem promises.

求 $f(x) = x^2$ 在 $[0,3]$ 上的平均值，并找出其达到平均值的点 $c$。

$$ f_{\text{avg}} = \frac{1}{3}\int_0^3 x^2\,dx = \frac{1}{3}\cdot \frac{27}{3} = 3. $$

在 $[0,3]$ 中解 $c^2 = 3$：$c = \sqrt{3} \approx 1.73$，位于区间内，符合定理的保证。

Find the average value of $f(x) = e^{-x}$ on $[0, 2]$.

$$ f_{\text{avg}} = \frac{1}{2 - 0}\int_0^2 e^{-x}\,dx = \frac{1}{2}\big[-e^{-x}\big]_0^2 = \frac{1}{2}\big(1 - e^{-2}\big) \approx \frac{1}{2}(0.8647) \approx 0.432. $$

Sanity check: $f$ runs from $f(0) = 1$ down to $f(2) = e^{-2} \approx 0.135$, so the average must lie strictly between $0.135$ and $1$, and $0.432$ does. Because $e^{-x}$ is convex, the average sits below the midpoint-of-endpoints value $\tfrac12(1 + 0.135) = 0.568$, which is a useful qualitative check.

求 $f(x) = e^{-x}$ 在 $[0, 2]$ 上的平均值。

$$ f_{\text{avg}} = \frac{1}{2}\int_0^2 e^{-x}\,dx = \frac{1}{2}\big(1 - e^{-2}\big) \approx 0.432. $$

验证：$f$ 从 $f(0)=1$ 降到 $f(2) \approx 0.135$，平均值必须严格介于两者之间，$0.432$ 满足。由于 $e^{-x}$ 是凸函数，平均值低于端点均值 $\tfrac12(1+0.135) = 0.568$，这是一个有用的定性检验。

A particle moves with velocity $v(t) = 3t^2$ on $[0, 2]$ (meters per second). Find its average velocity, and confirm it equals total displacement over elapsed time.

$$ v_{\text{avg}} = \frac{1}{2 - 0}\int_0^2 3t^2\,dt = \frac{1}{2}\big[t^3\big]_0^2 = \frac{1}{2}(8) = 4\,\text{m/s}. $$

The displacement is $\int_0^2 3t^2\,dt = 8\,\text{m}$ over $2\,\text{s}$, so displacement over time is $\tfrac{8}{2} = 4\,\text{m/s}$, matching. This is the physical content of the average value: it is the constant rate that would produce the same total accumulation over the same interval.

质点以速度 $v(t) = 3t^2$ 在 $[0, 2]$（米/秒）运动。求其平均速度，并验证其等于总位移除以时间。

$$ v_{\text{avg}} = \frac{1}{2}\int_0^2 3t^2\,dt = \frac{1}{2}(8) = 4\,\text{m/s}. $$

位移为 $\int_0^2 3t^2\,dt = 8\,\text{m}$，用时 $2\,\text{s}$，故位移除以时间 $= 4\,\text{m/s}$，与上一致。这正是平均值的物理意义：等效的常速率，能在相同区间内产生相同的总累积量。

Let $f$ be continuous on $[a,b]$. By the Extreme Value Theorem it attains a minimum $m$ and maximum $Mx$ there. Then $m \le f(x) \le Mx$, and integrating preserves the inequalities:

$$ m(b-a) \le \int_a^b f(x)\,dx \le Mx(b-a). $$

Dividing by $b-a$ shows the average value $f_{\text{avg}}$ lies between $m$ and $Mx$. Since $f$ is continuous, the Intermediate Value Theorem provides a point $c$ with $f(c) = f_{\text{avg}}$.

设 $f$ 在 $[a,b]$ 上连续。由极值定理（Extreme Value Theorem），$f$ 在此取得最小值 $m$ 和最大值 $Mx$。则 $m \le f(x) \le Mx$，积分保持不等式：

$$ m(b-a) \le \int_a^b f(x)\,dx \le Mx(b-a). $$

两边除以 $b-a$ 可知平均值 $f_{\text{avg}}$ 介于 $m$ 和 $Mx$ 之间。由于 $f$ 连续，介值定理（Intermediate Value Theorem）给出某点 $c$ 使得 $f(c) = f_{\text{avg}}$。

For what constant $k$ is $f(x) = kx$ on $[0,2]$ (and $0$ elsewhere) a valid density? Then find $P(0 \le X \le 1)$ and the mean.

Normalize: $\int_0^2 kx\,dx = k\cdot 2 = 1$, so $k = \tfrac12$. Then

$$ P(0 \le X \le 1) = \int_0^1 \tfrac12 x\,dx = \tfrac12\cdot \tfrac12 = \tfrac14. $$ $$ \mu = \int_0^2 x\cdot \tfrac12 x\,dx = \tfrac12\int_0^2 x^2\,dx = \tfrac12\cdot \tfrac{8}{3} = \tfrac{4}{3}. $$

对 $[0,2]$ 上的 $f(x) = kx$（其他处为 $0$），$k$ 取何值才是合法的密度函数？然后求 $P(0 \le X \le 1)$ 和均值。

归一化：$\int_0^2 kx\,dx = k\cdot 2 = 1$，故 $k = \tfrac12$。则：

$$ P(0 \le X \le 1) = \int_0^1 \tfrac12 x\,dx = \tfrac14. $$ $$ \mu = \int_0^2 x\cdot \tfrac12 x\,dx = \tfrac12\cdot \tfrac{8}{3} = \tfrac{4}{3}. $$

A call center's waiting time $X$ (minutes) follows the exponential density $f(x) = \tfrac{1}{\lambda}e^{-x/\lambda}$ for $x \ge 0$ with mean $\lambda = 5$. Verify it integrates to $1$, confirm the mean, and find $P(X > 10)$.

Normalization, with $\lambda = 5$:

$$ \int_0^\infty \tfrac{1}{5}e^{-x/5}\,dx = \big[-e^{-x/5}\big]_0^\infty = 0 - (-1) = 1. $$

The mean, by integration by parts ($u = x$, $dv = \tfrac15 e^{-x/5}dx$):

$$ \mu = \int_0^\infty x\cdot \tfrac{1}{5}e^{-x/5}\,dx = \big[-x e^{-x/5}\big]_0^\infty + \int_0^\infty e^{-x/5}\,dx = 0 + 5 = 5. $$

The tail probability that a caller waits more than $10$ minutes:

$$ P(X > 10) = \int_{10}^\infty \tfrac{1}{5}e^{-x/5}\,dx = \big[-e^{-x/5}\big]_{10}^\infty = e^{-2} \approx 0.135. $$

So about $13.5\%$ of callers wait beyond two mean wait times, a hallmark of the memoryless exponential.

某呼叫中心等待时间 $X$（分钟）服从指数分布，密度 $f(x) = \tfrac{1}{\lambda}e^{-x/\lambda}$（$x \ge 0$），均值 $\lambda = 5$。验证归一化，确认均值，并求 $P(X > 10)$。

归一化（$\lambda = 5$）：

$$ \int_0^\infty \tfrac{1}{5}e^{-x/5}\,dx = \big[-e^{-x/5}\big]_0^\infty = 1. $$

均值（分部积分，$u = x$，$dv = \tfrac15 e^{-x/5}dx$）：

$$ \mu = \int_0^\infty x\cdot \tfrac{1}{5}e^{-x/5}\,dx = 0 + 5 = 5. $$

等待超过 $10$ 分钟的尾部概率：

$$ P(X > 10) = \big[-e^{-x/5}\big]_{10}^\infty = e^{-2} \approx 0.135. $$

约 $13.5\%$ 的来电者等待时间超过两倍均值，这是指数分布无记忆性的体现。

For the density $f(x) = \tfrac12 x$ on $[0,2]$, find the median $m$, the value with half the probability on each side.

The median solves $\int_0^m \tfrac12 x\,dx = \tfrac12$:

$$ \tfrac12\cdot \tfrac{m^2}{2} = \tfrac12 \;\Longrightarrow\; \tfrac{m^2}{4} = \tfrac12 \;\Longrightarrow\; m^2 = 2 \;\Longrightarrow\; m = \sqrt{2} \approx 1.41. $$

Compare with the mean $\mu = \tfrac43 \approx 1.33$ from Worked Example 6.1. Here the median exceeds the mean. Because the density rises toward the right, the mass is pushed right, but the long-tail intuition (mean above median) applies to right-skewed densities; this density is left-light and right-heavy in a way that puts the median above the mean. The lesson is to compute, not to assume an ordering.

对密度 $f(x) = \tfrac12 x$（$[0,2]$ 上），求中位数 $m$（两侧各有一半概率的值）。

中位数满足 $\int_0^m \tfrac12 x\,dx = \tfrac12$：

$$ \tfrac{m^2}{4} = \tfrac12 \;\Longrightarrow\; m = \sqrt{2} \approx 1.41. $$

与例题 6.1 的均值 $\mu = \tfrac43 \approx 1.33$ 比较：中位数大于均值。由于密度向右递增，质量偏右，但"均值高于中位数"的直觉适用于右偏分布；本密度的中位数反而高于均值。教训是：要计算，不要假设两者的大小关系。

For a continuous random variable, the probability of any single exact value is zero. Formally, for any $a$,

$$ P(X = a) = \lim_{\varepsilon \to 0^+} \int_{a-\varepsilon}^{a+\varepsilon} f(x)\,dx = 0, $$

because the integral of a bounded $f$ over an interval of length $2\varepsilon$ is at most $(\max f)\,2\varepsilon \to 0$. This is why $P(a \le X \le b) = P(a < X < b)$ for continuous variables: including or excluding the endpoints changes the probability by zero. It also explains why $f(x)$ is a density, not a probability: $f(x)$ can exceed $1$, and only its integral over an interval carries probabilistic meaning. This is the precise sense in which continuous probability is area, not height.

对于连续随机变量，任何单个精确值的概率为零。形式上，对任意 $a$：

$$ P(X = a) = \lim_{\varepsilon \to 0^+} \int_{a-\varepsilon}^{a+\varepsilon} f(x)\,dx = 0, $$

因为有界 $f$ 在长度 $2\varepsilon$ 的区间上的积分至多为 $(\max f)\,2\varepsilon \to 0$。这就是为什么对连续变量有 $P(a \le X \le b) = P(a < X < b)$：端点的包含与否不改变概率。这也解释了为何 $f(x)$ 是密度而非概率：$f(x)$ 可以超过 $1$，只有 $f$ 在区间上的积分才有概率意义。这是连续概率"是面积而非高度"的精确含义。

Once the mean $\mu$ is known, the variance measures spread: $\sigma^2 = \int (x-\mu)^2 f(x)\,dx$. For the density $f(x) = \tfrac12 x$ on $[0,2]$ with $\mu = \tfrac43$:

$$ \sigma^2 = \int_0^2 \big(x - \tfrac43\big)^2 \tfrac12 x\,dx. $$

Expand using $E[X^2] = \int_0^2 x^2\cdot \tfrac12 x\,dx = \tfrac12\cdot \tfrac{16}{4} = 2$, then $\sigma^2 = E[X^2] - \mu^2 = 2 - \tfrac{16}{9} = \tfrac{2}{9}$.

知道均值 $\mu$ 后，方差衡量分散程度：$\sigma^2 = \int (x-\mu)^2 f(x)\,dx$。对密度 $f(x) = \tfrac12 x$（$[0,2]$ 上），$\mu = \tfrac43$：

$$ \sigma^2 = \int_0^2 \big(x - \tfrac43\big)^2 \tfrac12 x\,dx. $$

利用 $E[X^2] = \int_0^2 x^2\cdot \tfrac12 x\,dx = 2$，则 $\sigma^2 = E[X^2] - \mu^2 = 2 - \tfrac{16}{9} = \tfrac{2}{9}$。

The region under $y = x^2$ over $[0,1]$ is revolved about the line $x = 2$. Find the volume using Pappus, then note the alternative.

From Worked Example 4.1 the centroid is $\big(\tfrac34, \tfrac{3}{10}\big)$ and the area is $A = \tfrac13$. Revolving about the vertical line $x = 2$, the centroid is at distance $\bar{d} = 2 - \tfrac34 = \tfrac54$ from the axis. Pappus gives

$$ V = 2\pi\,\bar{d}\,A = 2\pi\cdot \tfrac54\cdot \tfrac13 = \frac{5\pi}{6}. $$

The shell-method integral $V = \int_0^1 2\pi(2 - x)x^2\,dx = 2\pi\int_0^1 (2x^2 - x^3)\,dx = 2\pi\big(\tfrac23 - \tfrac14\big) = 2\pi\cdot \tfrac{5}{12} = \tfrac{5\pi}{6}$ agrees, confirming the shortcut and tying Section 4 to solids of revolution.

将 $[0,1]$ 上 $y = x^2$ 曲线下的区域绕直线 $x = 2$ 旋转，用帕普斯定理（Pappus's theorem）求体积。

由例题 4.1，形心为 $\big(\tfrac34, \tfrac{3}{10}\big)$，面积 $A = \tfrac13$。绕竖直线 $x = 2$，形心到轴的距离 $\bar{d} = 2 - \tfrac34 = \tfrac54$。帕普斯定理给出：

$$ V = 2\pi\,\bar{d}\,A = 2\pi\cdot \tfrac54\cdot \tfrac13 = \frac{5\pi}{6}. $$

壳层法积分 $V = \int_0^1 2\pi(2-x)x^2\,dx = \tfrac{5\pi}{6}$ 与之吻合，验证了捷径，同时将第 4 节与旋转体联系起来。

A thin wire bent along $y = \sqrt{x}$ from $x=0$ to $x=4$ carries charge with linear density $\rho(x) = x$ per unit arc length. Set up the total charge using the slice-and-sum template.

The local quantity is (charge per unit length)(length of the slice). A slice of the curve has arc-length element $ds = \sqrt{1 + (y')^2}\,dx$, and $y' = \tfrac{1}{2\sqrt{x}}$, so $ds = \sqrt{1 + \tfrac{1}{4x}}\,dx$. The total charge is

$$ Q = \int_0^4 \rho(x)\,ds = \int_0^4 x\sqrt{1 + \frac{1}{4x}}\,dx = \int_0^4 x\cdot \frac{\sqrt{4x + 1}}{2\sqrt{x}}\,dx = \frac{1}{2}\int_0^4 \sqrt{x}\,\sqrt{4x+1}\,dx. $$

The point is not the messy antiderivative but the setup: identify the local rate (here $\rho\,ds$), pick the variable, and integrate. Every application in this unit is one instance of this single move.

一根细铁丝沿 $y = \sqrt{x}$ 弯曲（$x$ 从 $0$ 到 $4$），单位弧长（arc length）的线电荷密度为 $\rho(x) = x$。用切片求和模板建立总电荷的积分式。

局部量为（单位长度电荷）乘以（切片长度）。曲线的弧长（arc length）微元 $ds = \sqrt{1+(y')^2}\,dx$，$y' = \tfrac{1}{2\sqrt{x}}$，故 $ds = \sqrt{1+\tfrac{1}{4x}}\,dx$。总电荷：

$$ Q = \int_0^4 \rho(x)\,ds = \frac{1}{2}\int_0^4 \sqrt{x}\,\sqrt{4x+1}\,dx. $$

重点不在于混乱的原函数，而在于建立式子：识别局部速率（此处为 $\rho\,ds$），选取变量，积分。本单元的每个应用都是这一步骤的一个实例。

Pappus's theorem states that if a plane region of area $A$ is revolved about an external axis in its plane, the volume of the solid of revolution equals the area times the distance traveled by the centroid:

$$ V = 2\pi\,\bar{d}\,A, $$

where $\bar{d}$ is the distance from the centroid to the axis. For a disk of radius $r$ centered at distance $R$ from the axis ($R > r$), the centroid is the center, $\bar{d} = R$, and $A = \pi r^2$, giving the torus volume

$$ V = 2\pi R\cdot \pi r^2 = 2\pi^2 R r^2. $$

This converts a hard volume integral into a centroid lookup, tying Section 4 to the methods of solids of revolution.

帕普斯定理（Pappus's theorem）指出：面积为 $A$ 的平面区域绕平面内的外部轴旋转，所得旋转体的体积等于面积乘以形心走过的距离：

$$ V = 2\pi\,\bar{d}\,A, $$

其中 $\bar{d}$ 为形心（centroid）到轴的距离。对半径为 $r$、圆心距轴 $R$（$R>r$）的圆盘，$\bar{d} = R$，$A = \pi r^2$，给出圆环体（torus）的体积：

$$ V = 2\pi R\cdot \pi r^2 = 2\pi^2 R r^2. $$

这将困难的体积积分转化为形心查表，将第 4 节与旋转体方法联系起来。