积分与变化的累积

Differentiation 求导 found rates. Integration 积分 runs the idea in reverse: given a rate of change 变化率, it finds the accumulated change 累积变化. The key picture: the area 面积 between the graph of a rate function and the $x$-axis gives the total accumulation.

• If the rate is positive over an interval, the accumulated change is positive; if negative, negative. Area below the axis counts as negative.
• Simple regions 区域 (triangles, rectangles 矩形) can be found with geometry 几何.
• Units 单位: the area's unit is the rate's unit times the input's unit. A rate in vehicles-per-hour times hours gives vehicles.

When exact area is hard, approximate it with a Riemann sum 黎曼和 – split the interval into subintervals 子区间 and add up rectangle (or trapezoid 梯形) areas. The four standard estimates:

• Left Riemann sum – rectangle height from the left endpoint 端点 of each subinterval.
• Right Riemann sum – height from the right endpoint.
• Midpoint Riemann sum – height from the midpoint 中点.
• Trapezoidal sum 梯形法 – average the two endpoint heights (a trapezoid).

Subintervals may be uniform 等宽的 (equal width) or nonuniform 不等宽的 – read widths from the table.

Over- or underestimate? Judge from the behavior of the function: for an increasing function, a left sum underestimates and a right sum overestimates; a trapezoidal sum overestimates when the function is concave up and underestimates when concave down. Exam parts ask you to state which and why.

Worked example. A table gives $f(0)=3$, $f(2)=5$, $f(4)=8$, $f(6)=9$. Estimate $\int_0^6 f(x)\,dx$ with a right Riemann sum of three equal subintervals ($\Delta x=2$). Use the right endpoint of each strip:

Since $f$ is increasing, this right sum is an overestimate 高估; the left sum $2(3+5+8)=32$ would be an underestimate 低估.

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:

Meaning: about $54.5$ litres flow in during the first $10$ minutes. A common error is to use one width for every strip when the table steps are unequal.

As the subinterval widths shrink to zero, the Riemann sum approaches an exact value – the definite integral 定积分:

Here $\Delta x_i$ is the width of the $i$th subinterval and $x_i^{*}$ a point inside it. So a definite integral is the limit 极限 of a Riemann sum, and you should be able to translate each into the other.

A definite integral with a variable upper limit 上限 defines a new accumulation function 累积函数. The Fundamental Theorem of Calculus 微积分基本定理 (first part) says differentiation undoes this accumulation: if $f$ is continuous 连续, then

So if $g(x)=\int_a^x f(t)\,dt$, then $g'(x)=f(x)$ and $g''(x)=f'(x)$. This is the engine behind the very common “let $g(x)=\int_a^x f(t)\,dt$” questions.

When the upper limit is a function of $x$, the chain rule 链式法则 adds a factor:

Common mistakes. Forgetting the chain-rule factor $u'(x)$ when the upper limit is not plain $x$, and differentiating $f$ instead of evaluating it: $\dfrac{d}{dx}\displaystyle\int_a^x f(t)\,dt$ is $f(x)$, not $f'(x)$.

Because $g'(x)=f(x)$, everything from Unit 5 applies to an accumulation function using the graph of $f$:

• $g$ is increasing where $f>0$ and decreasing where $f<0$;
• $g$ has a local extremum 局部极值 where $f$ crosses zero (with a sign change 符号变化);
• $g$ is concave up 上凹 where $f$ is increasing; inflection points 拐点 of $g$ occur where $f$ has a local extremum.

To get a value of $g$, compute the signed area 有符号面积: $g(x)=\int_a^x f(t)\,dt$, adding areas above the axis and subtracting areas below.

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.

These properties simplify computation and appear constantly:

The last (splitting at an interior point 内点 $c$) lets you build a total integral from pieces read off a graph.

The second part of the Fundamental Theorem evaluates a definite integral using an antiderivative 原函数. If $F'=f$, then

So: find any antiderivative $F$, then subtract its values at the two limits. This is how most exact integrals are computed. It also gives the net change 净变化 view: $\int_a^b g'(t)\,dt = g(b)-g(a)$, so a starting value plus accumulated change gives a later value, e.g. $g(5)=g(0)+\int_0^5 g'(t)\,dt$.

Worked example. Evaluate $\int_1^3 (2x+1)\,dx$. An antiderivative is $F(x)=x^2+x$, so

Common mistake. Dropping the brackets: $9+3-1+1=12$ is wrong. Subtract the whole of $F(1)$.

An indefinite integral 不定积分 is the family of all antiderivatives, written with a constant of integration 积分常数:

Reverse each derivative rule to build the basic antiderivatives:

Common mistakes. Leaving out $+C$ on an indefinite integral; losing the minus sign in $\int\sin x\,dx=-\cos x+C$; and using the power rule on $x^{-1}$, which would divide by $0$: $\int\frac1x\,dx=\ln|x|+C$.

$u$-substitution 换元积分法 reverses the chain rule. Choose an inside function $u=g(x)$, so $du=g'(x)\,dx$, and rewrite the integral entirely in $u$:

Look for a function and its derivative 导数 both present. For a definite integral, either change the limits to $u$-values or convert back to $x$ before substituting the original limits.

Worked example. Evaluate $\int 2x\cos(x^2)\,dx$. The inside function is $u=x^2$, whose derivative $2x\,dx=du$ is present, so

Spotting that $2x$ is exactly $\dfrac{du}{dx}$ is the whole trick.

Worked example: definite integral. 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 $u=2$. Then

With the new limits there is no need to go back to $x$. Keeping the old limits $0$ and $1$ with $u$ is a common error.

Two algebraic set-up moves let more integrals fit the basic forms: polynomial long division 多项式长除法 when the top degree 次数 is $\ge$ the bottom degree of a rational function 有理函数, and completing the square 配方法 to turn a quadratic 二次 denominator 分母 into a form that integrates to an arctangent 反正切 or logarithm 对数.

Worked example: long division. Find $\displaystyle\int \frac{x^2+1}{x+1}\,dx$. The top degree ($2$) is at least the bottom degree ($1$), so divide first: $x^2+1=(x+1)(x-1)+2$. Then

Worked example: completing the square. Find $\displaystyle\int \frac{1}{x^2+2x+5}\,dx$. Complete the square: $x^2+2x+5=(x+1)^2+4$. With $u=x+1$ and $\displaystyle\int\frac{du}{u^2+a^2}=\frac1a\arctan\frac{u}{a}+C$ (here $a=2$):

Check any answer like this by differentiating it back.

A skill topic: match the integral to a method. Try a basic antiderivative first; look for a $u$-substitution (an inside function whose derivative is also present); use algebra (division, completing the square, splitting a fraction) to reshape the integrand 被积函数 into a standard form. Naming the structure first prevents wasted effort.

• For $n\ne-1$, use $\int x^n\,dx=\tfrac{x^{n+1}}{n+1}+C$ on an interval where the power rule 幂法则 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 (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.