Applications of Integration
AP Calculus BC Topic 8 9:59 English narration · English + 中文 subtitles burned in
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Here is a bumpy graph, and the area underneath it.
这是一条起伏的曲线，还有它下面的面积。
Now imagine pouring that area into a rectangle over the same interval, and letting it level off.
现在想象把这块面积倒进同一个区间上的一个矩形里， 让它自己找平。
The height it settles at is the average value of the function.
它稳定下来的那个高度，就是这个函数的平均值。
Same interval, same area, but one flat height instead of a wiggly one.
同一个区间、同样的面积，但高度是平的，不再是起伏的。
That single picture is where this whole unit starts.
整个单元就从这一张图开始。
Every application in this unit is the same move. Find a quantity you can slice thinly, write down the area or the volume of one slice, and integrate.
本单元的每一个应用都是同一个动作：找到一个可以切成薄片的量， 写下一片的面积或体积，然后积分。
Slice it thin, write one slice, and integrate.
切成薄片，写出一片，然后积分。
Let's begin.
让我们开始吧。
The average value of a function is the integral, divided by the width of the interval.
一个函数的平均值，就是它的积分除以区间的宽度。
Nothing more.
就这么简单。
It is the constant height a rectangle would need to have the same area as the region under the curve.
它是一个矩形要与曲线下方区域面积相等时所需的恒定高度。 做一道。
Work one. Take x squared on the interval from zero to three.
取 x 平方，区间从 0 到 3。
The integral is x cubed over three, evaluated from zero to three, which is nine.
积分是三分之 x 的三次方，从 0 到 3 取值，得到 9。
Divide by the width, which is three, and the average value is three.
除以宽度 3，平均值就是 3。
One warning. Do not confuse it with the average rate of change. Average value comes from an integral.
有一个提醒：不要把它和平均变化率搞混。
Average rate of change comes from a difference quotient.
平均值来自积分，平均变化率来自差商。
Unit four differentiated position to get velocity, and velocity to get acceleration.
第四单元对位置求导得到速度，对速度求导得到加速度。
Integration runs it backwards.
积分把这个过程倒过来。
Integrate acceleration and you get velocity. Integrate velocity and you get position.
对加速度积分得到速度，对速度积分得到位置。
And now the distinction that is worth marks on nearly every paper.
现在讲那个几乎每张卷子都值分的区别。
The integral of velocity is displacement, the net change in position, and it can be negative or even zero.
速度的积分是位移，也就是位置的净变化， 它可以是负的，甚至可以是零。
The integral of the size of velocity is total distance, which counts every metre travelled no matter which way.
速度大小的积分是总路程， 它把走过的每一米都算进去，不管方向。
To compute total distance you must split where velocity changes sign, and add the sizes of the pieces.
要算总路程，你必须在速度变号的地方拆开，再把各段的大小加起来。
The same idea applies far beyond motion.
同样的思想远远不止用于运动。
When a rate is given, water flowing in litres per minute, or sales in items per day, the definite integral of that rate gives the accumulated total, in the units of the rate times time.
当题目给出一个变化率时， 比如水以每分钟多少升流入，或者销量是每天多少件， 这个变化率的定积分就给出累积总量，单位是变化率的单位乘以时间。
Here is the setup that appears again and again. The amount at a later time equals the initial amount, plus the integral of rate in minus rate out.
有一个反复出现的题型：后来某时刻的量，等于初始量，加上"流入率减流出率"的积分。
Two things earn the marks. Do not forget the initial amount. And always interpret your answer in context, with units.
有两件事能拿到分：不要忘记初始量；以及永远要结合情境、带上单位来解释你的答案。
Now areas between two curves.
现在讲两条曲线之间的面积。
Do not memorise a formula. Draw one thin vertical strip instead.
不要去背公式，而是画出一条细细的竖条。
Its width is a tiny step in x, and its height is the upper curve minus the lower curve.
它的宽度是 x 上的一小步，高度是上面那条曲线减去下面那条曲线。
So the area of one strip is top minus bottom, times that tiny width.
所以一条竖条的面积就是"上减下"乘以那个很小的宽度。
Add up the strips, and that is the integral.
把所有竖条加起来，就是那个积分。
The limits are where the curves intersect, so set the two functions equal and solve for those first.
积分上下限就是两条曲线的交点，所以先令两个函数相等，把交点解出来。
And note this. Top minus bottom stays correct even when the region dips below the axis, because the subtraction takes care of the sign for you.
还要注意：即使区域落到横轴下方，"上减下"仍然是对的，因为这个减法已经替你处理好符号了。
Work the standard one.
来做这道标准题。
Find the area between y equals x, and y equals x squared.
求 y 等于 x 与 y 等于 x 平方之间的面积。
First the limits: set the two equal.
先求上下限：把两者设为相等。
x equals x squared, so x is zero or one.
x 等于 x 平方，所以 x 是零或一。
Those are your limits.
这就是你的上下限。
Now decide which is on top.
现在判断哪一条在上面。
Test any point between them, say a half: the line gives a half, the parabola gives a quarter, so the line is on top across the whole interval.
在两者之间任取 一点检验，比如二分之一：直线给出二分之一，抛物线给出四分之一，所以在整个区间上直线在上面。
Now write the integral, from zero to one, of x minus x squared.
现在写出积分，从零到一，被积函数是 x 减 x 平方。
Take the antiderivative term by term: x squared over two, minus x cubed over three.
逐项求原函数：x 平方比二，减 x 立方比三。
Evaluate at one, and the lower limit contributes nothing.
在一处取值，而下限没有贡献。
A half minus a third is one sixth.
二分之一减三分之一等于六分之一。
So the area is one sixth.
所以面积是六分之一。
Sometimes vertical strips are the wrong choice.
有时候竖条是错误的选择。
If a region's left and right boundaries are easier to write as x in terms of y, then turn the strip sideways. Use a thin horizontal strip instead.
如果一个区域的左右边界更容易写成"x 关于 y 的函数"， 那就把细条转过来横着用。
Its height is a tiny step in y, and its length is right minus left. The limits are now y values. Why bother?
用一条细细的横条，它的高度是 y 上的一小步， 长度是"右减左"。
Because choosing y can avoid splitting the region into several pieces.
这时上下限变成 y 的值。 为什么要这么做？
If a vertical strip would change which curve is on top halfway across, a horizontal strip often runs cleanly from one boundary to the other.
因为选择对 y 积分可以避免把区域拆成好几块。 如果一条竖条走到一半时上方的曲线换了，横条往往能从一边边界干净地贯穿到另一边。
And what if the curves cross more than twice?
那如果两条曲线交叉超过两次呢？
Then somewhere along the way, the top and bottom switch.
那么在中间某个地方，上下就换了位置。
If you integrate straight through, the two parts partly cancel, and you get the net signed area, not the real area.
如果你一口气积到底，两部分会部分抵消，你得到的是净的有符号面积，而不是真正的面积。
So split at each intersection.
所以要在每个交点处拆开。
On each piece, decide which curve is on top there, integrate top minus bottom for that piece, and add the pieces at the end.
在每一段上判断那里哪条曲线在上面， 对那一段积"上减下"，最后把各段加起来。
Written compactly, that is the integral of the size of the difference.
紧凑地写出来，就是"差的绝对值"的积分。
But on paper, show the split.
但在卷面上，要把拆分过程写出来。
Volumes work the same way, but now the slice is a thin slab.
体积的做法是一样的，只是这次切片变成了一块薄板。
Picture a solid sitting on a region in the plane, with every cross section perpendicular to the x axis being the same shape.
想象一个立体放在平面上的某个区域上，它每一个垂直于 x 轴的横截面都是同一种形状。
Take one thin slab. Its volume is the area of that cross section, times a tiny step in x.
取一块薄板，它的体积是那个横截面的面积，乘以 x 上的一小步。
So integrate the area.
所以对面积积分。
If the cross section is a square whose side is the distance between the two curves, the area is that distance squared.
如果横截面是一个正方形，边长等于两条曲线之间的距离，那么面积就是那个距离的平方。
For triangles and semicircles, just use the right area formula with the same distance as the base or the diameter.
对三角形和半圆，就用对应的面积公式，把同一个距离当作底或直径。
One motto covers every case. Integrate the cross sectional area.
有一句口诀涵盖所有情况：对横截面面积积分。
Now spin a region around an axis.
现在把一个区域绕一条轴旋转。
The solid you sweep out has circular cross sections, so the cross sections are discs, and the area of a disc is pi times the radius squared.
扫出来的立体有圆形的横截面， 所以横截面是圆盘，而圆盘的面积是 π 乘以半径的平方。
That gives the disc method. Pi, times the integral of the radius squared.
这就给出圆盘法： π 乘以半径平方的积分。
And the only real work is the radius.
真正要动脑的只有半径。
The radius is the distance from the curve to the axis of revolution.
半径是从曲线到旋转轴的距离。
When you revolve around the x axis, that distance is just the height of the curve.
当你绕 x 轴旋转时，这个距离就是曲线的高度。
Work one.
来做一道。
Take the region under y equals the square root of x, from zero to four, and spin it about the x axis.
取 y 等于根号 x 在零到四之间下方的区域，把它绕 x 轴旋转。
Draw one disc.
画出一个圆盘。
The radius is the height of the curve there, so the radius is the square root of x.
半径就是 曲线在那里的高度，所以半径是根号 x。
Now the disc method: pi times the integral of the radius squared.
现在用圆盘法：π 乘以半径平方的积分。
And here is the gift: the square root squares to just x.
好消息来了： 根号平方之后就只剩 x。
So the integral is pi times the integral of x, from zero to four.
所以积分是 π 乘以 x 从零到四的积分。
The antiderivative of x is x squared over two.
x 的原函数是 x 平方比二。
At four that is sixteen over two, which is eight, and zero contributes nothing.
在四处是十六比二，也就是八，而零没有贡献。
So the volume is eight pi.
所以体积是 八 π。
Always simplify inside the integral before you integrate.
在积分之前一定先把被积函数化简。
Exams almost always revolve around a line that is not an axis, something like y equals two.
考试几乎总是绕一条不是坐标轴的直线旋转，比如 y 等于 2。
Nothing about the method changes, but the radius adjusts.
方法本身没有任何变化， 但半径要作调整。
The radius is still the distance from the curve to the axis, so now it is the height of the curve, minus that line's height.
半径仍然是从曲线到旋转轴的距离，所以现在它是曲线的高度， 减去那条直线的高度。
Take the size of that, so the radius is never negative.
取它的绝对值，这样半径就永远不会是负的。
The habit that prevents every mistake here is simple. Sketch the region and the axis, then label the radius on your sketch before you write a single integral.
在这里能避免所有错误的习惯很简单：先画出区域和那条轴， 在图上把半径标出来，然后再动笔写积分。
But what if the region does not touch the axis?
但如果这个区域并不接触旋转轴呢？
Then spinning it leaves a hole down the middle, and each cross section is a ring, or a washer.
那么旋转之后中间会留下一个洞， 每个横截面就是一个圆环，也就是垫圈。
So subtract the inner disc from the outer disc.
所以要从外圆盘里减掉内圆盘。
The volume is pi, times the integral of the outer radius squared minus the inner radius squared.
体积等于 π 乘以"外半径平方减内半径平方"的积分。
One warning that costs marks every year. It is the difference of the squares, not the square of the difference.
有一个每年都在丢分的提醒：是平方之差，不是差的平方。
Those are not the same thing.
这两者不是一回事。
Find each radius separately as a distance to the axis, square each one, then subtract.
分别把每个半径当作到轴的距离求出来，各自平方，然后再相减。
Last idea, and it is BC only. How long is a curve?
最后一个概念，而且是 BC 独有的。
Zoom in on a tiny piece.
一条曲线有多长？
It is almost straight, so it is a tiny hypotenuse, with a small step across and a small step up.
把很小的一段放大来看。 它几乎是直的，所以它是一段很小的斜边，横向走了一小步，纵向也走了一小步。
Pythagoras gives its length.
勾股定理给出它的长度。
Now factor the small step in x out of the square root, and what is left inside is one plus the derivative squared.
现在把 x 上的那一小步从根号里提出来， 根号里剩下的就是 1 加上导数的平方。
Add up every tiny hypotenuse, and you get the arc length formula. The integral of the square root of one plus the derivative squared.
把每一段小斜边都加起来，就得到弧长公式： 根号下"1 加导数平方"的积分。
The same idea gives the distance a particle travels along a curved path.
同样的思想给出一个质点沿曲线路径走过的距离。
Work the standard one.
来做那道标准题。
Find the arc length of two thirds x to the power three halves, from zero to three. Differentiate first.
求三分之二乘以 x 的二分之三次方，从 0 到 3 的弧长。
The derivative is the square root of x.
先求导。 导数是根号 x。
Square it and you get x.
把它平方就得到 x。
So inside the square root we have one plus x, which is a perfect setup, because the ugly square root has vanished.
所以根号里面是 1 加 x， 这是一个非常好的设置，因为那个讨厌的根号消失了。
Integrate the square root of one plus x. That gives two thirds, times one plus x, to the power three halves.
对根号下"1 加 x"积分， 得到三分之二乘以"1 加 x"的二分之三次方。
Evaluate from zero to three. Two thirds times eight, minus two thirds times one, which is fourteen over three.
从 0 到 3 取值： 三分之二乘以 8，减去三分之二乘以 1，等于三分之十四。
Three marks students lose.
学生最常丢分的三个地方。
First, total distance needs the size of velocity, so split where the velocity changes sign.
第一，总路程要用速度的绝对值，所以要在速度变号的地方拆开。
Second, the washer method takes the difference of squares, never the square of the difference.
第二，垫圈法用的是平方之差，绝不是差的平方。
Third, when the axis is not a coordinate axis, sketch it first and label each radius before writing the integral.
第三，当旋转轴不是坐标轴时，先画图，把每个半径标出来，再写积分。