Unit A8: Antiderivatives and the Definite Integral第 A8 单元：原函数与定积分

A particle moving on a line has acceleration $a(t) = 6t$, initial velocity $v(0) = 4$, and initial position $s(0) = 1$. Find $s(t)$.

The velocity is an antiderivative of acceleration:

$$v(t) = \int 6t\,dt = 3t^{2} + C_{1}.$$

Imposing $v(0) = 4$ gives $C_{1} = 4$, so $v(t) = 3t^{2} + 4$. The position is an antiderivative of velocity:

$$s(t) = \int (3t^{2} + 4)\,dt = t^{3} + 4t + C_{2}.$$

Imposing $s(0) = 1$ gives $C_{2} = 1$, so $s(t) = t^{3} + 4t + 1$.

Notice the structure: each antidifferentiation introduced a fresh constant, and each initial condition consumed exactly one of them. Reversing a second-order rate of change therefore requires two pieces of initial data, which is why a falling-body problem needs both an initial height and an initial velocity.

一个在直线上运动的质点，加速度为 $a(t) = 6t$，初速度 $v(0) = 4$，初始位置 $s(0) = 1$。求 $s(t)$。

速度是加速度的原函数：

$$v(t) = \int 6t\,dt = 3t^{2} + C_{1}.$$

代入 $v(0) = 4$ 得 $C_{1} = 4$，所以 $v(t) = 3t^{2} + 4$。位置是速度的原函数：

$$s(t) = \int (3t^{2} + 4)\,dt = t^{3} + 4t + C_{2}.$$

代入 $s(0) = 1$ 得 $C_{2} = 1$，所以 $s(t) = t^{3} + 4t + 1$。

注意其中的结构：每做一次反求导都会引入一个新的常数，而每个初始条件恰好消去其中一个。因此逆转一个二阶变化率需要两条初始数据，这正是落体问题既需要初始高度又需要初始速度的原因。

Find $\displaystyle\int \frac{x^{2} + 3\sqrt{x} - 1}{x}\,dx$.

The power rule applies only to a sum of powers of $x$, so first split the quotient term by term and convert every root to a fractional exponent:

$$\frac{x^{2} + 3\sqrt{x} - 1}{x} = x + 3x^{-1/2} - x^{-1}.$$

Now antidifferentiate each piece. The middle term uses the power rule with $n = -\tfrac12$, and the last term uses the logarithm rule because its exponent is $-1$:

$$\int \big(x + 3x^{-1/2} - x^{-1}\big)\,dx = \frac{x^{2}}{2} + 3\cdot\frac{x^{1/2}}{1/2} - \ln|x| + C = \frac{x^{2}}{2} + 6\sqrt{x} - \ln|x| + C.$$

A quick differentiation check returns each original term, confirming the answer.

求 $\displaystyle\int \frac{x^{2} + 3\sqrt{x} - 1}{x}\,dx$。

幂法则只适用于 $x$ 的幂之和，所以先把分式逐项拆开，并把每个根号都写成分数指数：

$$\frac{x^{2} + 3\sqrt{x} - 1}{x} = x + 3x^{-1/2} - x^{-1}.$$

现在逐项反求导。中间一项用幂法则，取 $n = -\tfrac12$；最后一项因指数为 $-1$ 而要用对数法则：

$$\int \big(x + 3x^{-1/2} - x^{-1}\big)\,dx = \frac{x^{2}}{2} + 3\cdot\frac{x^{1/2}}{1/2} - \ln|x| + C = \frac{x^{2}}{2} + 6\sqrt{x} - \ln|x| + C.$$

快速求导验算会还原出每个原始项，从而确认答案无误。

Find the curve $y = F(x)$ whose slope at every point is $F'(x) = 3x^{2} - 2x$ and which passes through $(1, 4)$.

The general antiderivative is the whole family of solution curves:

$$F(x) = \int (3x^{2} - 2x)\,dx = x^{3} - x^{2} + C.$$

The point condition selects one member. Substituting $x = 1$, $y = 4$ gives $4 = 1 - 1 + C$, so $C = 4$ and

$$F(x) = x^{3} - x^{2} + 4.$$

Geometrically the constant $C$ slides the entire curve vertically, and the single point pins down which vertical translate to keep.

求曲线 $y = F(x)$，它在每一点的斜率为 $F'(x) = 3x^{2} - 2x$，且经过点 $(1, 4)$。

一般原函数就是整族解曲线：

$$F(x) = \int (3x^{2} - 2x)\,dx = x^{3} - x^{2} + C.$$

定点条件从中挑出一条。代入 $x = 1$、$y = 4$ 得 $4 = 1 - 1 + C$，所以 $C = 4$，于是

$$F(x) = x^{3} - x^{2} + 4.$$

从几何上看，常数 $C$ 把整条曲线上下平移，而这个定点确定了应保留哪一条竖直平移后的曲线。

Suppose $F$ and $G$ are both antiderivatives of $f$ on an interval $I$. Let $H = F - G$. Then for every $x$ in $I$,

$$H'(x) = F'(x) - G'(x) = f(x) - f(x) = 0.$$

A function with zero derivative on an interval is constant: this is a corollary of the Mean Value Theorem, since for any $a < b$ in $I$ there is a point $c$ with $H(b) - H(a) = H'(c)(b - a) = 0$. Hence $H(x) = C$ for some constant, so $F(x) = G(x) + C$. The interval hypothesis is essential: on the disconnected domain of $1/x$ the constant may differ on each piece.

设 $F$ 与 $G$ 都是 $f$ 在区间 $I$ 上的原函数。令 $H = F - G$。则对 $I$ 上每个 $x$，

$$H'(x) = F'(x) - G'(x) = f(x) - f(x) = 0.$$

在区间上导数恒为零的函数必为常数：这是中值定理（Mean Value Theorem）的一个推论，因为对 $I$ 中任意 $a < b$，都存在一点 $c$ 使 $H(b) - H(a) = H'(c)(b - a) = 0$。于是 $H(x) = C$ 为某常数，所以 $F(x) = G(x) + C$。区间这个前提至关重要：在 $1/x$ 那种不连通的定义域上，常数在每一片上可能不同。

Estimate the area under $f(x) = x^{2}$ on $[0,1]$ using $n$ equal subintervals and right endpoints, then take $n \to \infty$.

Here $\Delta x = 1/n$ and $x_{i} = i/n$, so

$$R_{n} = \sum_{i=1}^{n} \left(\frac{i}{n}\right)^{2}\frac{1}{n} = \frac{1}{n^{3}}\sum_{i=1}^{n} i^{2} = \frac{1}{n^{3}}\cdot\frac{n(n+1)(2n+1)}{6}.$$

Expanding, $R_{n} = \dfrac{(n+1)(2n+1)}{6n^{2}} = \dfrac{2n^{2} + 3n + 1}{6n^{2}}$, which tends to $\dfrac{2}{6} = \dfrac{1}{3}$ as $n \to \infty$. The exact area is $\tfrac{1}{3}$.

用 $n$ 个等长子区间和右端点估计 $f(x) = x^{2}$ 在 $[0,1]$ 上下方的面积，然后令 $n \to \infty$。

这里 $\Delta x = 1/n$，$x_{i} = i/n$，所以

$$R_{n} = \sum_{i=1}^{n} \left(\frac{i}{n}\right)^{2}\frac{1}{n} = \frac{1}{n^{3}}\sum_{i=1}^{n} i^{2} = \frac{1}{n^{3}}\cdot\frac{n(n+1)(2n+1)}{6}.$$

展开得 $R_{n} = \dfrac{(n+1)(2n+1)}{6n^{2}} = \dfrac{2n^{2} + 3n + 1}{6n^{2}}$，当 $n \to \infty$ 时趋于 $\dfrac{2}{6} = \dfrac{1}{3}$。精确面积为 $\tfrac{1}{3}$。

Compute $\displaystyle\int_{1}^{3} x^{2}\,dx$ as a limit of right Riemann sums.

Now the left endpoint is $a = 1$, so $\Delta x = \dfrac{3-1}{n} = \dfrac{2}{n}$ and the right endpoints are $x_{i} = 1 + i\,\Delta x = 1 + \dfrac{2i}{n}$. Squaring,

$$f(x_{i}) = \left(1 + \frac{2i}{n}\right)^{2} = 1 + \frac{4i}{n} + \frac{4i^{2}}{n^{2}}.$$

Multiply by $\Delta x = 2/n$ and sum, using $\sum 1 = n$, $\sum i = \tfrac{n(n+1)}{2}$, and $\sum i^{2} = \tfrac{n(n+1)(2n+1)}{6}$:

$$R_{n} = \frac{2}{n}\sum_{i=1}^{n}\left(1 + \frac{4i}{n} + \frac{4i^{2}}{n^{2}}\right) = \frac{2}{n}\left(n + \frac{4}{n}\cdot\frac{n(n+1)}{2} + \frac{4}{n^{2}}\cdot\frac{n(n+1)(2n+1)}{6}\right).$$

Simplify the three pieces to $2 + \dfrac{4(n+1)}{n} + \dfrac{4(n+1)(2n+1)}{3n^{2}}$. As $n \to \infty$ these tend to $2 + 4 + \dfrac{8}{3} = \dfrac{26}{3}$. The FTC check agrees: $\big[\tfrac{x^{3}}{3}\big]_{1}^{3} = 9 - \tfrac13 = \tfrac{26}{3}$.

把 $\displaystyle\int_{1}^{3} x^{2}\,dx$ 写成右黎曼和（RRS）的极限来计算。

现在左端点为 $a = 1$，故 $\Delta x = \dfrac{3-1}{n} = \dfrac{2}{n}$，右端点为 $x_{i} = 1 + i\,\Delta x = 1 + \dfrac{2i}{n}$。平方得

$$f(x_{i}) = \left(1 + \frac{2i}{n}\right)^{2} = 1 + \frac{4i}{n} + \frac{4i^{2}}{n^{2}}.$$

乘以 $\Delta x = 2/n$ 再求和，用到 $\sum 1 = n$、$\sum i = \tfrac{n(n+1)}{2}$ 和 $\sum i^{2} = \tfrac{n(n+1)(2n+1)}{6}$：

$$R_{n} = \frac{2}{n}\sum_{i=1}^{n}\left(1 + \frac{4i}{n} + \frac{4i^{2}}{n^{2}}\right) = \frac{2}{n}\left(n + \frac{4}{n}\cdot\frac{n(n+1)}{2} + \frac{4}{n^{2}}\cdot\frac{n(n+1)(2n+1)}{6}\right).$$

把三部分化简为 $2 + \dfrac{4(n+1)}{n} + \dfrac{4(n+1)(2n+1)}{3n^{2}}$。当 $n \to \infty$ 时它们趋于 $2 + 4 + \dfrac{8}{3} = \dfrac{26}{3}$。用微积分基本定理（FTC）验算一致：$\big[\tfrac{x^{3}}{3}\big]_{1}^{3} = 9 - \tfrac13 = \tfrac{26}{3}$。

For $f(x) = x^{2}$ on $[0,1]$ with $n = 4$ equal strips, find $L_{4}$ and $R_{4}$ and state what they tell you about the true area.

Here $\Delta x = \tfrac14$ and the partition points are $0, \tfrac14, \tfrac12, \tfrac34, 1$. The left sum samples the left endpoints:

$$L_{4} = \tfrac14\Big(0^{2} + \big(\tfrac14\big)^{2} + \big(\tfrac12\big)^{2} + \big(\tfrac34\big)^{2}\Big) = \tfrac14\cdot\tfrac{14}{16} = \tfrac{7}{32} \approx 0.219.$$

The right sum samples the right endpoints:

$$R_{4} = \tfrac14\Big(\big(\tfrac14\big)^{2} + \big(\tfrac12\big)^{2} + \big(\tfrac34\big)^{2} + 1^{2}\Big) = \tfrac14\cdot\tfrac{30}{16} = \tfrac{15}{32} \approx 0.469.$$

Because $f$ is increasing, $L_{4} \le \text{area} \le R_{4}$, so the exact value $\tfrac13 \approx 0.333$ is trapped between them. The gap $R_{4} - L_{4} = \tfrac14\big(1^{2} - 0^{2}\big) = \tfrac14$ shrinks like $\tfrac{1}{n}$, which is why refining the partition forces both sums to the same limit.

对 $[0,1]$ 上的 $f(x) = x^{2}$，取 $n = 4$ 个等宽竖条，求 $L_{4}$ 与 $R_{4}$，并说明它们对真实面积意味着什么。

这里 $\Delta x = \tfrac14$，分点为 $0, \tfrac14, \tfrac12, \tfrac34, 1$。左和取各子区间的左端点：

$$L_{4} = \tfrac14\Big(0^{2} + \big(\tfrac14\big)^{2} + \big(\tfrac12\big)^{2} + \big(\tfrac34\big)^{2}\Big) = \tfrac14\cdot\tfrac{14}{16} = \tfrac{7}{32} \approx 0.219.$$

右和取各子区间的右端点：

$$R_{4} = \tfrac14\Big(\big(\tfrac14\big)^{2} + \big(\tfrac12\big)^{2} + \big(\tfrac34\big)^{2} + 1^{2}\Big) = \tfrac14\cdot\tfrac{30}{16} = \tfrac{15}{32} \approx 0.469.$$

由于 $f$ 递增，$L_{4} \le \text{area} \le R_{4}$，所以精确值 $\tfrac13 \approx 0.333$ 被夹在两者之间。差值 $R_{4} - L_{4} = \tfrac14\big(1^{2} - 0^{2}\big) = \tfrac14$ 以 $\tfrac{1}{n}$ 的量级缩小，这正是细化分割能迫使两个和趋于同一极限的原因。

The closed form $\sum_{i=1}^{n} i^{2} = \tfrac{n(n+1)(2n+1)}{6}$ is the engine behind the $\int x^2$ computations, and it follows from a telescoping trick. Start from the algebraic identity

$$(i+1)^{3} - i^{3} = 3i^{2} + 3i + 1.$$

Sum both sides from $i = 1$ to $n$. The left side telescopes, since each $i^3$ cancels against the next term, leaving only the endpoints:

$$\sum_{i=1}^{n}\big[(i+1)^{3} - i^{3}\big] = (n+1)^{3} - 1.$$

The right side splits using known sums, $\sum i = \tfrac{n(n+1)}{2}$ and $\sum 1 = n$:

$$(n+1)^{3} - 1 = 3\sum_{i=1}^{n} i^{2} + 3\cdot\frac{n(n+1)}{2} + n.$$

Solve for $\sum i^{2}$. Expanding $(n+1)^3 - 1 = n^3 + 3n^2 + 3n$ and isolating the unknown sum gives, after combining terms, $\sum_{i=1}^{n} i^{2} = \tfrac{n(n+1)(2n+1)}{6}$. The same telescoping idea with the fourth power yields the formula for $\sum i^3$.

闭式 $\sum_{i=1}^{n} i^{2} = \tfrac{n(n+1)(2n+1)}{6}$ 是各种 $\int x^2$ 计算背后的引擎，它来自一个裂项相消（telescoping）的技巧。从代数恒等式出发：

$$(i+1)^{3} - i^{3} = 3i^{2} + 3i + 1.$$

对 $i = 1$ 到 $n$ 把两边求和。左边裂项相消，因为每个 $i^3$ 都与下一项抵消，只剩下两端：

$$\sum_{i=1}^{n}\big[(i+1)^{3} - i^{3}\big] = (n+1)^{3} - 1.$$

右边用已知的和拆开，$\sum i = \tfrac{n(n+1)}{2}$ 和 $\sum 1 = n$：

$$(n+1)^{3} - 1 = 3\sum_{i=1}^{n} i^{2} + 3\cdot\frac{n(n+1)}{2} + n.$$

解出 $\sum i^{2}$。展开 $(n+1)^3 - 1 = n^3 + 3n^2 + 3n$ 并把待求和孤立出来，合并同类项后得 $\sum_{i=1}^{n} i^{2} = \tfrac{n(n+1)(2n+1)}{6}$。对四次幂用同样的裂项思路即可得到 $\sum i^3$ 的公式。

Evaluate $\displaystyle\int_{-1}^{2} (2x)\,dx$ by interpreting it as signed area.

The graph of $y = 2x$ is a line through the origin. On $[-1,0]$ it lies below the axis, forming a triangle of area $\tfrac{1}{2}(1)(2) = 1$ counted negatively. On $[0,2]$ it lies above the axis, forming a triangle of area $\tfrac{1}{2}(2)(4) = 4$ counted positively. The signed total is $-1 + 4 = 3$.

This matches the antiderivative computation $\big[x^{2}\big]_{-1}^{2} = 4 - 1 = 3$.

把 $\displaystyle\int_{-1}^{2} (2x)\,dx$ 理解成带符号面积来求值。

$y = 2x$ 的图像是过原点的一条直线。在 $[-1,0]$ 上它位于轴下方，构成面积 $\tfrac{1}{2}(1)(2) = 1$ 的三角形，记为负；在 $[0,2]$ 上它位于轴上方，构成面积 $\tfrac{1}{2}(2)(4) = 4$ 的三角形，记为正。带符号的总和为 $-1 + 4 = 3$。

这与用原函数计算的结果一致：$\big[x^{2}\big]_{-1}^{2} = 4 - 1 = 3$。

Evaluate $\displaystyle\int_{-2}^{2} x^{3}\,dx$ and contrast it with the geometric area between the curve and the axis.

The integrand is odd: $f(-x) = -f(x)$. On $[-2,0]$ the curve lies below the axis and on $[0,2]$ it lies above, and the two regions are mirror images. Their signed contributions cancel exactly, so

$$\int_{-2}^{2} x^{3}\,dx = \Big[\frac{x^{4}}{4}\Big]_{-2}^{2} = \frac{16}{4} - \frac{16}{4} = 0.$$

This is the signed area. The geometric (unsigned) area between the curve and the axis is not zero; it equals $2\int_{0}^{2} x^{3}\,dx = 2\cdot 4 = 8$. The lesson is that a definite integral reports the net of positive and negative pieces, and an odd integrand over a symmetric interval always nets to zero.

求 $\displaystyle\int_{-2}^{2} x^{3}\,dx$，并与曲线和坐标轴之间的几何面积作对比。

被积函数是奇函数：$f(-x) = -f(x)$。在 $[-2,0]$ 上曲线在轴下方，在 $[0,2]$ 上在轴上方，两块区域互为镜像。它们带符号的贡献恰好抵消，于是

$$\int_{-2}^{2} x^{3}\,dx = \Big[\frac{x^{4}}{4}\Big]_{-2}^{2} = \frac{16}{4} - \frac{16}{4} = 0.$$

这是带符号面积。曲线与坐标轴之间的几何（无符号）面积并不为零；它等于 $2\int_{0}^{2} x^{3}\,dx = 2\cdot 4 = 8$。要点是：定积分给出的是正负两部分的净值，而奇函数在对称区间上的积分总是净得零。

Given $\displaystyle\int_{0}^{4} f = 10$ and $\displaystyle\int_{0}^{4} g = -3$, find $\displaystyle\int_{4}^{0}\big(2f - 5g\big)\,dx$.

First use linearity on the forward interval $[0,4]$:

$$\int_{0}^{4}\big(2f - 5g\big)\,dx = 2\int_{0}^{4} f - 5\int_{0}^{4} g = 2(10) - 5(-3) = 20 + 15 = 35.$$

The requested integral runs from $4$ to $0$, which reverses orientation and flips the sign:

$$\int_{4}^{0}\big(2f - 5g\big)\,dx = -\int_{0}^{4}\big(2f - 5g\big)\,dx = -35.$$

No formula for $f$ or $g$ was needed; the algebra of integrals did all the work.

已知 $\displaystyle\int_{0}^{4} f = 10$ 和 $\displaystyle\int_{0}^{4} g = -3$，求 $\displaystyle\int_{4}^{0}\big(2f - 5g\big)\,dx$。

先在正向区间 $[0,4]$ 上用线性性：

$$\int_{0}^{4}\big(2f - 5g\big)\,dx = 2\int_{0}^{4} f - 5\int_{0}^{4} g = 2(10) - 5(-3) = 20 + 15 = 35.$$

所求积分是从 $4$ 到 $0$，这翻转了方向，符号也随之改变：

$$\int_{4}^{0}\big(2f - 5g\big)\,dx = -\int_{0}^{4}\big(2f - 5g\big)\,dx = -35.$$

整个过程不需要 $f$ 或 $g$ 的具体表达式；积分的运算法则完成了全部工作。

Two order properties follow directly from the fact that a limit of sums preserves inequalities. If $f(x) \le g(x)$ for all $x$ in $[a,b]$, then every Riemann sum of $f$ is term-by-term no larger than the matching sum of $g$, since $f(x_i^{*})\,\Delta x \le g(x_i^{*})\,\Delta x$. Passing to the limit preserves the inequality:

$$f \le g \text{ on } [a,b] \;\Longrightarrow\; \int_{a}^{b} f\,dx \le \int_{a}^{b} g\,dx.$$

Specializing to constant bounds $m \le f(x) \le M$ gives the estimate

$$m(b-a) \le \int_{a}^{b} f(x)\,dx \le M(b-a),$$

because $\int_a^b m\,dx = m(b-a)$. For example, on $[0,1]$ the function $f(x) = \tfrac{1}{1+x^{2}}$ satisfies $\tfrac12 \le f(x) \le 1$, so its integral lies between $\tfrac12$ and $1$ without computing it (the exact value is $\tfrac{\pi}{4} \approx 0.785$, comfortably inside). These bounds are the workhorse behind convergence and error estimates throughout Calculus II.

有两条序性质直接来自“和的极限保持不等号”这一事实。若对 $[a,b]$ 中所有 $x$ 都有 $f(x) \le g(x)$，则 $f$ 的每个黎曼和都逐项不大于 $g$ 的对应和，因为 $f(x_i^{*})\,\Delta x \le g(x_i^{*})\,\Delta x$。取极限保持不等号：

$$f \le g \text{ on } [a,b] \;\Longrightarrow\; \int_{a}^{b} f\,dx \le \int_{a}^{b} g\,dx.$$

特殊化为常数界 $m \le f(x) \le M$ 时，得到估计

$$m(b-a) \le \int_{a}^{b} f(x)\,dx \le M(b-a),$$

因为 $\int_a^b m\,dx = m(b-a)$。例如在 $[0,1]$ 上，函数 $f(x) = \tfrac{1}{1+x^{2}}$ 满足 $\tfrac12 \le f(x) \le 1$，所以不必计算就知道它的积分介于 $\tfrac12$ 与 $1$ 之间（精确值为 $\tfrac{\pi}{4} \approx 0.785$，确实落在区间内部）。这些界是微积分 II 中各种收敛性（convergence）与误差估计背后的主力工具。

Let $g(x) = \displaystyle\int_{2}^{x^{2}} \sqrt{1 + t^{3}}\,dt$. Find $g'(x)$.

With $f(t) = \sqrt{1 + t^{3}}$ and upper limit $u(x) = x^{2}$, the chain rule variant gives

$$g'(x) = f\big(x^{2}\big)\cdot \frac{d}{dx}\big(x^{2}\big) = \sqrt{1 + (x^{2})^{3}}\cdot 2x = 2x\sqrt{1 + x^{6}}.$$

设 $g(x) = \displaystyle\int_{2}^{x^{2}} \sqrt{1 + t^{3}}\,dt$。求 $g'(x)$。

取 $f(t) = \sqrt{1 + t^{3}}$、上限 $u(x) = x^{2}$，链式法则变体给出

$$g'(x) = f\big(x^{2}\big)\cdot \frac{d}{dx}\big(x^{2}\big) = \sqrt{1 + (x^{2})^{3}}\cdot 2x = 2x\sqrt{1 + x^{6}}.$$

Let $\displaystyle h(x) = \int_{x}^{x^{2}} \sin(t^{2})\,dt$. Find $h'(x)$.

When both limits move, first split the integral at any fixed constant $c$ to isolate two accumulation functions, using additivity and the sign flip on the lower piece:

$$h(x) = \int_{x}^{c} \sin(t^{2})\,dt + \int_{c}^{x^{2}} \sin(t^{2})\,dt = -\int_{c}^{x} \sin(t^{2})\,dt + \int_{c}^{x^{2}} \sin(t^{2})\,dt.$$

Differentiate each with the chain-rule variant of FTC Part 1:

$$h'(x) = -\sin(x^{2}) + \sin\big((x^{2})^{2}\big)\cdot 2x = 2x\sin(x^{4}) - \sin(x^{2}).$$

The general pattern is $\dfrac{d}{dx}\int_{a(x)}^{b(x)} f = f(b(x))b'(x) - f(a(x))a'(x)$: top limit minus bottom limit, each weighted by its own derivative.

设 $\displaystyle h(x) = \int_{x}^{x^{2}} \sin(t^{2})\,dt$。求 $h'(x)$。

当上下限都变动时，先在任一固定常数 $c$ 处把积分拆开，用可加性以及下半段的符号翻转，分离出两个累积函数：

$$h(x) = \int_{x}^{c} \sin(t^{2})\,dt + \int_{c}^{x^{2}} \sin(t^{2})\,dt = -\int_{c}^{x} \sin(t^{2})\,dt + \int_{c}^{x^{2}} \sin(t^{2})\,dt.$$

用 FTC 第一部分的链式法则变体分别求导：

$$h'(x) = -\sin(x^{2}) + \sin\big((x^{2})^{2}\big)\cdot 2x = 2x\sin(x^{4}) - \sin(x^{2}).$$

一般规律是 $\dfrac{d}{dx}\int_{a(x)}^{b(x)} f = f(b(x))b'(x) - f(a(x))a'(x)$：上限项减下限项，每一项各自乘以自己的导数。

Let $\displaystyle g(x) = \int_{0}^{x} (t^{2} - 4)\,dt$. Without evaluating the integral, find where $g$ has a local maximum or minimum on $[0,4]$.

By FTC Part 1, $g'(x) = x^{2} - 4$, which is zero at $x = 2$ (the value $x = -2$ is outside the interval). Since $g'$ changes from negative to positive there, $g'(x) < 0$ on $(0,2)$ and $g'(x) > 0$ on $(2,4)$, the accumulation function decreases then increases, so $x = 2$ is a local minimum.

This is the power of Part 1: the sign of the integrand controls the monotonicity of the area-so-far function, so you can analyze $g$ entirely from $f = g'$ without ever computing $g$ in closed form.

设 $\displaystyle g(x) = \int_{0}^{x} (t^{2} - 4)\,dt$。在不计算积分的情况下，求 $g$ 在 $[0,4]$ 上取得极大值或极小值的位置。

由 FTC 第一部分，$g'(x) = x^{2} - 4$，它在 $x = 2$ 处为零（$x = -2$ 落在区间外）。由于 $g'$ 在该处由负变正，即 $g'(x) < 0$ 在 $(0,2)$ 上、$g'(x) > 0$ 在 $(2,4)$ 上，累积函数先减后增，所以 $x = 2$ 是极小值点。

这正是第一部分的威力：被积函数的符号控制着“已累积面积”函数的单调性，所以你完全可以从 $f = g'$ 来分析 $g$，而无需求出 $g$ 的闭式表达式。

Fix $x$ and form the difference quotient of $g(x) = \int_{a}^{x} f(t)\,dt$:

$$\frac{g(x+h) - g(x)}{h} = \frac{1}{h}\int_{x}^{x+h} f(t)\,dt.$$

Because $f$ is continuous on $[x, x+h]$, the Extreme Value Theorem gives a minimum value $m$ and maximum value $M$ on that interval, so $mh \le \int_{x}^{x+h} f(t)\,dt \le Mh$ (taking $h > 0$). Dividing by $h$,

$$m \le \frac{g(x+h) - g(x)}{h} \le M.$$

As $h \to 0$, both $m$ and $M$ approach $f(x)$ by continuity (squeeze). Therefore the difference quotient has limit $f(x)$, which is exactly $g'(x) = f(x)$.

固定 $x$，写出 $g(x) = \int_{a}^{x} f(t)\,dt$ 的差商：

$$\frac{g(x+h) - g(x)}{h} = \frac{1}{h}\int_{x}^{x+h} f(t)\,dt.$$

由于 $f$ 在 $[x, x+h]$ 上连续，极值定理（EVT）保证它在该区间上取得最小值 $m$ 与最大值 $M$，于是 $mh \le \int_{x}^{x+h} f(t)\,dt \le Mh$（取 $h > 0$）。两边除以 $h$：

$$m \le \frac{g(x+h) - g(x)}{h} \le M.$$

当 $h \to 0$ 时，由连续性，$m$ 与 $M$ 都趋于 $f(x)$（夹逼）。因此差商的极限为 $f(x)$，这正是 $g'(x) = f(x)$。

Evaluate $\displaystyle\int_{0}^{\pi/2} \big(\sin x + 3x^{2}\big)\,dx$.

An antiderivative is $F(x) = -\cos x + x^{3}$. By FTC Part 2,

$$\int_{0}^{\pi/2}\!\big(\sin x + 3x^{2}\big)\,dx = \Big[-\cos x + x^{3}\Big]_{0}^{\pi/2}.$$

At the upper limit $-\cos(\pi/2) + (\pi/2)^{3} = 0 + \pi^{3}/8$. At the lower limit $-\cos 0 + 0 = -1$. The difference is

$$\frac{\pi^{3}}{8} - (-1) = \frac{\pi^{3}}{8} + 1.$$

求 $\displaystyle\int_{0}^{\pi/2} \big(\sin x + 3x^{2}\big)\,dx$。

一个原函数是 $F(x) = -\cos x + x^{3}$。由 FTC 第二部分，

$$\int_{0}^{\pi/2}\!\big(\sin x + 3x^{2}\big)\,dx = \Big[-\cos x + x^{3}\Big]_{0}^{\pi/2}.$$

在上限处 $-\cos(\pi/2) + (\pi/2)^{3} = 0 + \pi^{3}/8$；在下限处 $-\cos 0 + 0 = -1$。两者之差为

$$\frac{\pi^{3}}{8} - (-1) = \frac{\pi^{3}}{8} + 1.$$

Water flows into a tank at a rate $r(t) = 12 - 3t$ liters per minute for $0 \le t \le 4$. Find the net change in volume over this interval.

Volume is the antiderivative of flow rate, so the net change is the integral of the rate (the net change theorem). Using FTC Part 2 with $V(t) = 12t - \tfrac{3}{2}t^{2}$,

$$\Delta V = \int_{0}^{4}(12 - 3t)\,dt = \Big[12t - \tfrac{3}{2}t^{2}\Big]_{0}^{4} = \big(48 - 24\big) - 0 = 24 \text{ liters}.$$

The rate is positive on $[0,4)$ and reaches zero at $t = 4$, so the tank is filling throughout and the net change equals the actual volume added. Had the rate gone negative, the integral would have reported inflow minus outflow.

水以 $r(t) = 12 - 3t$ 升每分钟的速率流入水箱，$0 \le t \le 4$。求这段区间内体积的净变化。

体积是流速的原函数，所以净变化就是速率的积分（净变化定理）。用 FTC 第二部分，取 $V(t) = 12t - \tfrac{3}{2}t^{2}$，

$$\Delta V = \int_{0}^{4}(12 - 3t)\,dt = \Big[12t - \tfrac{3}{2}t^{2}\Big]_{0}^{4} = \big(48 - 24\big) - 0 = 24 \text{ liters}.$$

速率在 $[0,4)$ 上为正、在 $t = 4$ 处变为零，所以水箱始终在注水，净变化等于实际注入的体积。若速率曾变负，积分给出的就是流入减去流出的结果。

Evaluate $\displaystyle\int_{1}^{e} \frac{2 + \ln x}{x}\,dx$.

Split the integrand: $\dfrac{2}{x} + \dfrac{\ln x}{x}$. The first term antidifferentiates to $2\ln x$. For the second, recognize $\dfrac{\ln x}{x}$ as $u\,du$ with $u = \ln x$, whose antiderivative is $\tfrac12(\ln x)^{2}$. So an antiderivative is $F(x) = 2\ln x + \tfrac12(\ln x)^{2}$, valid on all of $[1,e]$ since $x > 0$ there. Then

$$\int_{1}^{e} \frac{2 + \ln x}{x}\,dx = \Big[2\ln x + \tfrac12(\ln x)^{2}\Big]_{1}^{e} = \big(2 + \tfrac12\big) - \big(0 + 0\big) = \frac{5}{2}.$$

求 $\displaystyle\int_{1}^{e} \frac{2 + \ln x}{x}\,dx$。

把被积函数拆开：$\dfrac{2}{x} + \dfrac{\ln x}{x}$。第一项反求导得 $2\ln x$。对第二项，把 $\dfrac{\ln x}{x}$ 识别为 $u\,du$，其中 $u = \ln x$，其原函数为 $\tfrac12(\ln x)^{2}$。于是一个原函数是 $F(x) = 2\ln x + \tfrac12(\ln x)^{2}$，由于 $[1,e]$ 上处处 $x > 0$，它在整个区间上有效。于是

$$\int_{1}^{e} \frac{2 + \ln x}{x}\,dx = \Big[2\ln x + \tfrac12(\ln x)^{2}\Big]_{1}^{e} = \big(2 + \tfrac12\big) - \big(0 + 0\big) = \frac{5}{2}.$$

Let $F$ be any antiderivative of the continuous function $f$, and let $g(x) = \int_{a}^{x} f(t)\,dt$. By FTC Part 1, $g' = f = F'$, so $g$ and $F$ differ by a constant: $g(x) = F(x) + C$. Evaluating at $x = a$ gives $g(a) = \int_{a}^{a} f = 0$, hence $C = -F(a)$. Now evaluate at $x = b$:

$$\int_{a}^{b} f(t)\,dt = g(b) = F(b) + C = F(b) - F(a).$$

This shows that any antiderivative may be used, since the additive constant cancels in the subtraction.

设 $F$ 是连续函数 $f$ 的任意一个原函数，令 $g(x) = \int_{a}^{x} f(t)\,dt$。由 FTC 第一部分，$g' = f = F'$，所以 $g$ 与 $F$ 相差一个常数：$g(x) = F(x) + C$。在 $x = a$ 处取值得 $g(a) = \int_{a}^{a} f = 0$，于是 $C = -F(a)$。再在 $x = b$ 处取值：

$$\int_{a}^{b} f(t)\,dt = g(b) = F(b) + C = F(b) - F(a).$$

这表明任何一个原函数都可使用，因为相加常数会在作差时抵消。

Evaluate $\displaystyle\int 2x\,(x^{2} + 1)^{5}\,dx$.

Let $u = x^{2} + 1$, so $du = 2x\,dx$. The integral becomes

$$\int u^{5}\,du = \frac{u^{6}}{6} + C = \frac{(x^{2}+1)^{6}}{6} + C.$$

求 $\displaystyle\int 2x\,(x^{2} + 1)^{5}\,dx$。

令 $u = x^{2} + 1$，则 $du = 2x\,dx$。积分变为

$$\int u^{5}\,du = \frac{u^{6}}{6} + C = \frac{(x^{2}+1)^{6}}{6} + C.$$

Evaluate $\displaystyle\int_{0}^{1} x\,e^{x^{2}}\,dx$.

Let $u = x^{2}$, so $du = 2x\,dx$ and $x\,dx = \tfrac{1}{2}\,du$. When $x = 0$, $u = 0$; when $x = 1$, $u = 1$. Thus

$$\int_{0}^{1} x\,e^{x^{2}}\,dx = \frac{1}{2}\int_{0}^{1} e^{u}\,du = \frac{1}{2}\big[e^{u}\big]_{0}^{1} = \frac{1}{2}(e - 1).$$

求 $\displaystyle\int_{0}^{1} x\,e^{x^{2}}\,dx$。

令 $u = x^{2}$，则 $du = 2x\,dx$，$x\,dx = \tfrac{1}{2}\,du$。当 $x = 0$ 时 $u = 0$；当 $x = 1$ 时 $u = 1$。于是

$$\int_{0}^{1} x\,e^{x^{2}}\,dx = \frac{1}{2}\int_{0}^{1} e^{u}\,du = \frac{1}{2}\big[e^{u}\big]_{0}^{1} = \frac{1}{2}(e - 1).$$

Evaluate $\displaystyle\int x\sqrt{x + 1}\,dx$.

Set $u = x + 1$, so $du = dx$ and $x = u - 1$. Here the substitution does not absorb the stray $x$ outright, so solve for it and rewrite the whole integrand in $u$:

$$\int x\sqrt{x+1}\,dx = \int (u - 1)\sqrt{u}\,du = \int \big(u^{3/2} - u^{1/2}\big)\,du.$$

Now the power rule finishes it:

$$= \frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2} + C = \frac{2}{5}(x+1)^{5/2} - \frac{2}{3}(x+1)^{3/2} + C.$$

This back-substitution technique handles many integrands where the derivative of the inner function is not already present as a factor.

求 $\displaystyle\int x\sqrt{x + 1}\,dx$。

令 $u = x + 1$，则 $du = dx$，$x = u - 1$。这里换元并不能直接把多出来的 $x$ 吸收掉，所以把它反解出来，并把整个被积函数都改写成 $u$：

$$\int x\sqrt{x+1}\,dx = \int (u - 1)\sqrt{u}\,du = \int \big(u^{3/2} - u^{1/2}\big)\,du.$$

接下来用幂法则收尾：

$$= \frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2} + C = \frac{2}{5}(x+1)^{5/2} - \frac{2}{3}(x+1)^{3/2} + C.$$

这种回代技巧能处理许多被积函数中内层函数的导数并未现成地作为因子出现的情形。

Evaluate $\displaystyle\int_{0}^{\pi/4} \tan x\,\sec^{2} x\,dx$.

The factor $\sec^{2} x$ is exactly the derivative of $\tan x$, so let $u = \tan x$, giving $du = \sec^{2} x\,dx$. Convert the limits: $x = 0 \Rightarrow u = 0$ and $x = \pi/4 \Rightarrow u = 1$. Then

$$\int_{0}^{\pi/4} \tan x\,\sec^{2} x\,dx = \int_{0}^{1} u\,du = \Big[\frac{u^{2}}{2}\Big]_{0}^{1} = \frac{1}{2}.$$

Recognizing a function paired with its own derivative is the single most useful pattern-matching skill for substitution.

求 $\displaystyle\int_{0}^{\pi/4} \tan x\,\sec^{2} x\,dx$。

因子 $\sec^{2} x$ 恰好是 $\tan x$ 的导数，所以令 $u = \tan x$，得 $du = \sec^{2} x\,dx$。替换上下限：$x = 0 \Rightarrow u = 0$，$x = \pi/4 \Rightarrow u = 1$。于是

$$\int_{0}^{\pi/4} \tan x\,\sec^{2} x\,dx = \int_{0}^{1} u\,du = \Big[\frac{u^{2}}{2}\Big]_{0}^{1} = \frac{1}{2}.$$

认出“一个函数与它自己的导数成对出现”是换元法中最有用的一项模式识别技能。

Substitution is the chain rule integrated. Suppose $F$ is an antiderivative of $f$, so $F' = f$. By the chain rule,

$$\frac{d}{dx}\,F\big(g(x)\big) = F'\big(g(x)\big)\,g'(x) = f\big(g(x)\big)\,g'(x).$$

Reading this equation as an antiderivative statement says exactly

$$\int f\big(g(x)\big)\,g'(x)\,dx = F\big(g(x)\big) + C = \int f(u)\,du \Big|_{u = g(x)}.$$

For the definite version, apply FTC Part 2 to $F(g(x))$ on $[a,b]$: the result is $F(g(b)) - F(g(a))$, which is precisely $\int_{g(a)}^{g(b)} f(u)\,du$. This is why the limits travel through $g$. The informal manipulation $du = g'(x)\,dx$ is a faithful shorthand for this chain-rule bookkeeping, not a separate rule.

换元法就是把链式法则积分起来。设 $F$ 是 $f$ 的原函数，即 $F' = f$。由链式法则，

$$\frac{d}{dx}\,F\big(g(x)\big) = F'\big(g(x)\big)\,g'(x) = f\big(g(x)\big)\,g'(x).$$

把这个等式当作一句关于原函数的陈述来读，恰好就是

$$\int f\big(g(x)\big)\,g'(x)\,dx = F\big(g(x)\big) + C = \int f(u)\,du \Big|_{u = g(x)}.$$

对定积分版本，把 FTC 第二部分用到 $[a,b]$ 上的 $F(g(x))$：结果是 $F(g(b)) - F(g(a))$，这正是 $\int_{g(a)}^{g(b)} f(u)\,du$。这就是上下限要经过 $g$ 变换的原因。那条非正式的写法 $du = g'(x)\,dx$ 是对这套链式法则记账的忠实简写，而不是另一条独立的法则。

Find the average value of $f(x) = x^{2}$ on $[0,3]$.

By the average value formula,

$$f_{\text{avg}} = \frac{1}{3 - 0}\int_{0}^{3} x^{2}\,dx = \frac{1}{3}\left[\frac{x^{3}}{3}\right]_{0}^{3} = \frac{1}{3}\cdot\frac{27}{3} = 3.$$

The Mean Value Theorem for integrals guarantees a point $c$ in $[0,3]$ with $f(c) = 3$, namely $c = \sqrt{3}$.

求 $f(x) = x^{2}$ 在 $[0,3]$ 上的平均值。

由平均值公式，

$$f_{\text{avg}} = \frac{1}{3 - 0}\int_{0}^{3} x^{2}\,dx = \frac{1}{3}\left[\frac{x^{3}}{3}\right]_{0}^{3} = \frac{1}{3}\cdot\frac{27}{3} = 3.$$

积分中值定理（Mean Value Theorem）保证在 $[0,3]$ 中存在一点 $c$ 使 $f(c) = 3$，即 $c = \sqrt{3}$。

Find the area of the region between $y = x$ and $y = x^{2}$ on $[0,1]$, the prototype of every Calculus II application.

On $[0,1]$ the line lies above the parabola, since $x \ge x^{2}$ there. A thin vertical strip at position $x$ has height $x - x^{2}$ and width $dx$, so its area is $(x - x^{2})\,dx$. Summing the strips and taking the limit is the integral

$$A = \int_{0}^{1}\big(x - x^{2}\big)\,dx = \Big[\frac{x^{2}}{2} - \frac{x^{3}}{3}\Big]_{0}^{1} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}.$$

The recipe slice, approximate, sum, take the limit is identical to the Riemann-sum construction of Section 2; only the height of the strip changes from problem to problem. Volumes, arc length, and work all follow this same template.

求 $[0,1]$ 上 $y = x$ 与 $y = x^{2}$ 之间区域的面积——这是微积分 II 中每一类应用的原型。

在 $[0,1]$ 上直线位于抛物线上方，因为这里 $x \ge x^{2}$。位于 $x$ 处的一条细竖条高 $x - x^{2}$、宽 $dx$，所以其面积为 $(x - x^{2})\,dx$。把这些竖条相加再取极限就是积分

$$A = \int_{0}^{1}\big(x - x^{2}\big)\,dx = \Big[\frac{x^{2}}{2} - \frac{x^{3}}{3}\Big]_{0}^{1} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}.$$

“切片、近似、求和、取极限”这套配方与第 2 节的黎曼和构造完全一致；从一道题到另一道题，变的只是竖条的高度。体积、弧长和功都沿用同一个模板。

The average value formula is more than a definition; for continuous $f$ the average is actually attained. Claim. If $f$ is continuous on $[a,b]$, there exists $c$ in $[a,b]$ with

$$f(c) = \frac{1}{b-a}\int_{a}^{b} f(x)\,dx.$$

Proof. By the Extreme Value Theorem, $f$ attains a minimum $m$ and a maximum $M$ on $[a,b]$. The bounding property from Section 3 gives $m(b-a) \le \int_a^b f \le M(b-a)$, so the average value $A = \tfrac{1}{b-a}\int_a^b f$ satisfies $m \le A \le M$. Since $f$ is continuous and takes both the values $m$ and $M$, the Intermediate Value Theorem guarantees some $c$ in $[a,b]$ with $f(c) = A$. This $c$ is exactly the point where a rectangle of height $f(c)$ over $[a,b]$ has the same area as the region under $f$.

平均值公式不只是一个定义；对连续的 $f$，这个平均值是真正能取到的。命题。若 $f$ 在 $[a,b]$ 上连续，则在 $[a,b]$ 中存在 $c$ 使

$$f(c) = \frac{1}{b-a}\int_{a}^{b} f(x)\,dx.$$

证明。由极值定理（EVT），$f$ 在 $[a,b]$ 上取得最小值 $m$ 与最大值 $M$。第 3 节的估界性质给出 $m(b-a) \le \int_a^b f \le M(b-a)$，所以平均值 $A = \tfrac{1}{b-a}\int_a^b f$ 满足 $m \le A \le M$。由于 $f$ 连续且取到 $m$ 与 $M$ 两个值，介值定理（IVT）保证在 $[a,b]$ 中存在某个 $c$ 使 $f(c) = A$。这个 $c$ 恰好是这样一点：在 $[a,b]$ 上以 $f(c)$ 为高的矩形与 $f$ 下方的区域面积相等。