积分与变化的累积

Where a derivative 导数 measures a rate, an integral 积分 measures an accumulation 累积 – a total built up from a rate. If a rate of change acts over an interval, the area 面积 between its graph and the axis gives the net accumulated change. This "area = total change" idea is the foundation of integral calculus 积分学.

A Riemann sum 黎曼和 estimates the area under a curve by adding the areas of thin rectangles 矩形. Split $[a,b]$ into subintervals 子区间 and use the function's height at the left endpoint, right endpoint, or midpoint 中点 of each. A trapezoidal sum 梯形法 uses trapezoids 梯形 instead, averaging the two endpoint heights – usually more accurate. With more, thinner rectangles the estimate improves.

Exam skill: be able to compute left, right, midpoint, and trapezoidal estimates from a table or graph, and state whether each over- or under-estimates based on whether the function is increasing/decreasing or concave up/down.

Worked example: trapezoidal sum, unequal widths. Water flows into a tank at rate $R(t)$ litres per minute.

$t$ (minutes)	$0$	$2$	$5$	$10$
$R(t)$ (litres/min)	$4$	$6$	$7$	$3$

Estimate $\int_0^{10} R(t)\,dt$ with a trapezoidal sum. The widths are $2$, $3$ and $5$, so each trapezoid needs its own width:

A left Riemann sum on the same table is $2(4)+3(6)+5(7)=61$ litres. Using one width for every strip when the table steps are unequal is a common error.

Writing a Riemann sum with summation notation 求和记号 $\sum_{k=1}^{n} f(x_k)\,\Delta x$ and letting the number of rectangles grow without bound gives the exact area – the definite integral 定积分:

The integral is the limit of Riemann sums; $a$ and $b$ are the limits of integration 积分限.

An accumulation function 累积函数 $g(x)=\int_a^x f(t)\,dt$ gives the accumulated area from $a$ up to $x$. The Fundamental Theorem of Calculus (FTC) 微积分基本定理 says its derivative is the integrand 被积函数:

Differentiation and integration are inverse operations. With a variable upper limit 上限 and the chain rule 链式法则, $\dfrac{d}{dx}\int_a^{u(x)} f(t)\,dt=f(u(x))\,u'(x)$.

Common mistakes. Forgetting the chain-rule factor: $\dfrac{d}{dx}\displaystyle\int_1^{x^2}\cos t\,dt=\cos(x^2)\cdot 2x$, not $\cos(x^2)$. Differentiating $f$ instead of evaluating it: the derivative of $\int_a^x f(t)\,dt$ is $f(x)$, not $f'(x)$.

Because $g'(x)=f(x)$, the graph of $f$ tells you everything about $g$: $g$ increases where $f>0$, decreases where $f<0$, has extrema where $f$ crosses zero, and is concave up where $f$ is increasing. Reading these connections off a graph of $f$ is a classic free-response task.

Common mistake. Adding an area below the axis instead of subtracting it. Suppose $f$ lies above the axis on $[0,3]$ with area $6$ and below it on $[3,5]$ with area $2$. Then $g(5)=\int_0^5 f(t)\,dt=6-2=4$, and $g$ has its maximum at $x=3$, where $f$ changes from positive to negative. The total area, $8$, answers a different question.

Definite integrals obey useful rules: reversing the limits negates the value ($\int_b^a=-\int_a^b$), an integral over a zero-width interval is $0$, they add over adjacent intervals 在相邻区间上相加 ($\int_a^c=\int_a^b+\int_b^c$), and constants factor out. Use these to combine or split given integral values.

The evaluation form of the FTC computes a definite integral from an antiderivative 原函数 $F$ (where $F'=f$):

So integrating a rate of change over $[a,b]$ gives the net change 净变化 in the quantity – the single most-used result in the course.

Worked example. $\displaystyle\int_1^3 (2x+1)\,dx$: an antiderivative is $F(x)=x^2+x$, so the value is $F(3)-F(1)=12-2=10$.

An indefinite integral 不定积分 $\int f(x)\,dx=F(x)+C$ is the family of all antiderivatives (hence the constant of integration 积分常数 $C$). Reverse each derivative rule: the power rule 幂法则 becomes $\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C$ (for $n\neq-1$), with $\int \frac1x\,dx=\ln|x|+C$, and the antiderivatives of $e^x$, $\sin x$, $\cos x$, and $\sec^2 x$ come straight from their derivatives.

Common mistakes. Leaving out $+C$; losing the minus sign in $\int\sin x\,dx=-\cos x+C$; and using the power rule on $x^{-1}$, which would divide by $0$.

u-substitution 换元积分 reverses the chain rule: choose $u=g(x)$ so that $g'(x)$ also appears, turning $\int f(g(x))g'(x)\,dx$ into $\int f(u)\,du$. Remember to convert $dx$ to $du$ and, for a definite integral, either change the limits to $u$-values or convert back to $x$ at the end.

Worked example. Evaluate $\displaystyle\int_0^1 x\,(x^2+1)^3\,dx$. Let $u=x^2+1$, so $du=2x\,dx$ and $x\,dx=\tfrac12\,du$. Change the limits: $x=0$ gives $u=1$, and $x=1$ gives $u=2$:

When a rational 有理 integrand is “top-heavy” (numerator degree $\ge$ denominator degree), long division 长除法 rewrites it as a polynomial 多项式 plus a proper fraction 真分式 you can integrate. Completing the square 配方法 in a denominator 分母 turns it into a form like $u^2+a^2$, leading to an arctangent 反正切 antiderivative $\frac1a\arctan\frac{u}{a}+C$.

Worked example: long division. Since $x^2+1=(x+1)(x-1)+2$,

Worked example: completing the square. Since $x^2+2x+5=(x+1)^2+4$, take $u=x+1$ and $a=2$:

Integration by parts 分部积分 reverses the product rule:

Choose $u$ to simplify when differentiated and $dv$ to be easy to integrate (the LIATE guide: logs, inverse-trig, algebraic, trig, exponential). It handles products like $\int x e^x\,dx$ and $\int x\ln x\,dx$, sometimes applied twice.

Worked example. For $\int x e^x\,dx$, choose $u=x$ ($du=dx$) and $dv=e^x\,dx$ ($v=e^x$):

Worked example: definite. Evaluate $\displaystyle\int_1^{e}\ln x\,dx$. There is no obvious product, so take $u=\ln x$ ($du=\tfrac1x\,dx$) and $dv=dx$ ($v=x$):

Apply the limits to the $uv$ part too; dropping them there is a frequent slip.

Partial fractions 部分分式 split a rational function with a factorable denominator into a sum of simpler fractions:

each of which integrates to a logarithm 对数. This technique is essential for the logistic differential equation 逻辑斯谛微分方程 in the next unit.

Worked example. Find $\displaystyle\int\frac{x+7}{(x-1)(x+3)}\,dx$. Write $\dfrac{x+7}{(x-1)(x+3)}=\dfrac{A}{x-1}+\dfrac{B}{x+3}$ and multiply through by the denominator:

Put $x=1$: $8=4A$, so $A=2$. Put $x=-3$: $4=-4B$, so $B=-1$. Then

Check by recombining: $2(x+3)-(x-1)=x+7$.

Common mistakes. Splitting a top-heavy fraction without dividing first (when the top's degree is at least the bottom's, do long division before partial fractions), and writing $\int\frac{1}{x+3}\,dx$ with the wrong sign or as $\ln(x+3)$ without the absolute value.

An improper integral 反常积分 has an infinite limit of integration or an infinite discontinuity 无穷间断点 in the integrand. Evaluate it as a limit: $\int_a^\infty f\,dx=\lim_{b\to\infty}\int_a^b f\,dx$. If the limit is a finite number the integral converges 收敛; otherwise it diverges 发散. Split at every interior singularity 奇点, and split an integral over the whole real line at a finite point. Each resulting one-sided improper integral must converge separately; cancellation of infinite parts is not convergence.

Worked example. $\displaystyle\int_1^\infty \frac{1}{x^2}\,dx=\lim_{b\to\infty}\left[-\frac1x\right]_1^b=\lim_{b\to\infty}\left(1-\frac1b\right)=1$, so it converges to $1$. By contrast $\int_1^\infty \frac1x\,dx$ gives $\lim_{b\to\infty}\ln b=\infty$ and diverges – the same integrand-shape can go either way.

Worked example: infinite discontinuity. $\dfrac{1}{\sqrt{x}}$ is unbounded at $x=0$, so

which converges. An integral like this looks ordinary, so check the integrand for a vertical asymptote 竖直渐近线 on $[a,b]$ before using $F(b)-F(a)$ directly.

Common mistake. Missing a vertical asymptote inside the interval. Blindly writing

gives a negative answer for a positive integrand. The integrand is unbounded at $x=0$: split there, and $\int_0^1\frac{1}{x^2}\,dx$ diverges, so the whole integral diverges.

The BC exam expects you to recognize which method fits: basic rules, substitution (a chain-rule pattern), by parts (a product), partial fractions (a factorable rational), or long division/completing the square. Being able to look at an integral and pick the right tool quickly is itself a tested skill.

• For $n\ne-1$, use $\int x^n\,dx=\tfrac{x^{n+1}}{n+1}+C$ on an interval where the power rule applies. For $n=-1$, use $\ln|x|+C$. Do not forget $+C$.
• The Fundamental Theorem links the two: $\int_a^b f'(x)\,dx=f(b)-f(a)$, and $\tfrac{d}{dx}\int_a^x f(t)\,dt=f(x)$.
• Approximate a definite integral with Riemann sums or the trapezoidal rule from a table of values.
• A definite integral is a signed area 有符号面积 (below the axis counts negative); split at sign changes for total area.
• Use u-substitution and remember to change the limits (or back-substitute) accordingly.