课程： 15.10 曲线下区域

章节大纲

选择节常规

折叠展开
常规

全部折叠展开全部
- 选择活动新闻通告
  
  新闻通告讨论区

曲线下的区域

Introduction
::导言

Suppose a person is driving on the highway. The velocity of the car in meters per second can be modeled by a quadratic for the 1st 8 seconds of acceleration , $\begin{align*}v(t)=t^2\end{align*}$ . How far has the car traveled in those 8 seconds?
::假设有人在高速公路上驾驶。每秒每米的汽车速度可以用加速度1至8秒的四方形模型模拟, v(t)=t2。这8秒的汽车行驶有多远?

lesson content

Area Under a Curve
::曲线下的区域

The area under a curve refers to the area between a curve and the $\begin{align*}x\end{align*}$ -axis. This area could be entirely above the $\begin{align*}x\end{align*}$ -axis, entirely below the $\begin{align*}x\end{align*}$ -axis, or a combination of above and below the $\begin{align*}x\end{align*}$ -axis. In calculus, the area under a curve is a visual representation of an integral. Often the area under a curve can be interpreted as the accumulated amount of whatever the function is modeling.
::曲线下的区域指曲线和 X 轴之间的区域。这个区域可以完全高于 x 轴, 完全低于 x 轴, 或X 轴下方的组合。在微积分中, 曲线下的区域是一个整体的直观表示。曲线下的区域通常可以被解释为任何函数的累积量。

For instance, suppose a curve represents the velocity of a moving object over 4 hours. The amount of time in hours the object moves is the $\begin{align*}x\end{align*}$ variable, and the velocity in miles per hour is the $\begin{align*}y\end{align*}$ variable. The area under the curve represents the total distance traveled over 4 hours. Two additional examples are shown in the following video:
::例如,假设曲线代表移动对象在4小时内的速度。物体移动的时数是 x 变量,每小时的速度是 y 变量。曲线下的区域代表总距离超过 4 小时。以下视频中还展示了两个例子:

When the area under a curve is a simple shape, such as the area under a straight line, the area can be calculated using geometry. However, if the area under a curve is not a simple shape, then the area can either be approximated using rectangles or trapezoids, or the area can be calculated directly using integration methods taught in c alculus. The rest of this section will focus on approximating the area under a curve using rectangles.
::当曲线下的区域是一个简单的形状时,例如直线下的区域,可以使用几何法来计算区域。但是,如果曲线下的区域不是一个简单的形状,那么,可以使用矩形或三角形来比较该区域,或者可以直接使用微积分中教授的集成法来计算该区域。本节的其余部分将侧重于使用矩形来接近曲线下的区域。

As mentioned, the area under a curve can be approximated with rectangles equally spaced under a curve as shown below. This approximation method for t he area under a curve is called Riemann Sums.
::如前所述,曲线下的区域可以与以下曲线下相同间距的矩形相近。对于曲线下的区域,这种近似法称为Riemann Sums。

lesson content

For consistency, you can choose whether the boxes should hit the curve on the left-hand corner, the right-hand corner, the maximum value, or the minimum value. The blue approximation uses right-handed boxes. The red approximation assigns the height of the box to be the minimum value of the function in each subinterval. The green approximation assigns the height of the box to be the maximum value of the function in each subinterval. The yellow approximation uses left-handed boxes.
::为了一致性,您可以选择框是否应该打中左侧角的曲线、右手角、最大值或最小值。蓝色近似值使用右手边的框。红色近近似值将框的高度指定为每个次间距函数的最小值。绿色近近似值将框的高度指定为每个次间距函数的最大值。黄色近近似值使用左手边的框。

For each of these examples, the area of each rectangle is calculated using the equation $\begin{align*}\text{Area}= b\cdot h\end{align*}$ . Rectangles above the $\begin{align*}x\end{align*}$ -axis will have positive area because the height (or $\begin{align*}y\end{align*}$ -value) is positive. Rectangles below the $\begin{align*}x\end{align*}$ -axis will have negative area because the height (or $\begin{align*}y\end{align*}$ -value) is nega tive.
::对于其中的每一个例子,每个矩形的区域都使用方程式Area=bh计算。x轴上方的矩形将具有正区域,因为高度(或y-value)为正区域。x轴下方的矩形将具有负区域,因为高度(或y-value)为负区域。

The more boxes you use, the narrower the boxes will be, and thus the more accurate the approximation will be to the actual area. In fact, the limit of the area approximation as the number of boxes increases to infinity is the precise area under the curve.
::您使用的盒子越多, 盒子会越窄, 因此近似会越精确到实际区域。事实上, 当框数增加到无限时, 区域近似值的限度是曲线下的确切区域。

Relationship Between an Integral and the Area Under a Curve
::整体关系和曲线下 " 区域 " 之间的关系

\begin{aligned} \int_{1}^{\infty} f (x) = lim_{n \to \infty} \sum_{i = 1}^{n} (Area of each box) \end{aligned}

$\begin{align*}\int_{1}^{\infty}{f(x)}=\lim_{n \to \infty}\sum \limits_{i=1}^n (\text{Area of each box})\end{align*}$
::1f(x)=limni=1n(每个框的区域)

In general, the area under a curve using Riemann Sums can be calculated by adding the areas of all the rectangles where the bases are equally spaced.
::一般而言,使用Riemann Sums的曲线下区域可以通过增加所有矩形的区域来计算,而这些矩形的基座在其中同样有间距。

The area under a curve is calculated by
::曲线下的区域由

\begin{aligned} Area = \sum_{i = 1}^{n} f (c_{i}) \cdot Δ x, \end{aligned}

$\begin{align*}\text{Area}=\sum_{i=1}^{n}{f(c_i) \cdot \Delta x },\end{align*}$
::区域=1nf(ci)x,

where $\begin{align*}c_i\end{align*}$ is the point of the box that hits the curve, $\begin{align*}f(c_i)\end{align*}$ is the function value at that point, and $\begin{align*}\Delta x\end{align*}$ is the width of the base of each rectangle.
::此处的 ci 是击中曲线的框点, f( ci) 是该点的函数值, \\ x 是每个矩形底部的宽度。

For a curve defined on the interval $\begin{align*}[a,b]\end{align*}$ ,
::对于在间隔 [a,b] 上定义的曲线,

\begin{aligned} Δ x = \frac{b - a}{n}, \end{aligned}

$\begin{align*}\Delta x = \frac{b-a}{n},\end{align*}$
::x=b -an, x=b -an, x=b -an, x=b -an, x=b -an, x=b -an, x=b -an, x=b -an,

where $\begin{align*}n\end{align*}$ is the number of rectangles or boxes.
::n 是矩形数或框数。

An example using this formula can be seen in the following video:
::使用此公式的例子可见于以下视频:

Play, Learn, and Explore Area Under a Curve with :
::在曲线下玩、学习和探索区域 :

Examples
::实例

Example 1
::例1

Use four right-handed boxes to approximate the area between 1 and 9 of the function $\begin{align*}f(x)=\frac{1}{2}x-2\end{align*}$ .
::使用四个右手框以接近函数 f( x) = 12x-2 1 和 9 之间的区域。

Solution:
::解决方案 :

Step 1: For the interval $\begin{align*}[1,9]\end{align*}$ with 4 rectangles, the width of the base of each rectangle is
::第1步:对于有4个矩形的间隔[1,9]和4个矩形,每个矩形的基底宽度是

\begin{aligned} Δ x = \frac{9 - 1}{4} = \frac{8}{4} = 2. \end{aligned}

$\begin{align*}\Delta x = \frac{9-1}{4}=\frac{8}{4}=2.\end{align*}$
::x=9 -14=84=2

Since the rectangles hit the curve on the right-hand corner, the 1st intersection point will be 2 units (the base width) from the start of the interval, 1. Thus, the 1st $\begin{align*}x\end{align*}$ -value the rectangle hits the curve is at $\begin{align*}x=3\end{align*}$ . The next $\begin{align*}x\end{align*}$ -values are 5, 7, and 9.
::由于矩形击中右手角的曲线,第一个交叉点将是从间隔开始的2个单位(基宽),1。因此,矩形击中曲线的第1个X值是 x=3。下一个X值是5、7和9。

Step 2: Calculate each corresponding function value for these $\begin{align*}x\end{align*}$ -values. Each function value is the height of its corresponding rectangle.
::第2步:为这些 x 值计算相应的函数值。每个函数值是相应的矩形的高度。

\begin{aligned} f (3) & = \frac{1}{2} (3) - 2 = - \frac{1}{2} \\ f (5) & = \frac{1}{2} (5) - 2 = \frac{1}{2} \\ f (7) & = \frac{1}{2} (7) - 2 = \frac{3}{2} \\ f (9) & = \frac{1}{2} (9) - 2 = \frac{5}{2} \end{aligned}

$\begin{align*}f(3)&=\frac{1}{2}(3)-2=-\frac{1}{2} \\ f(5)&=\frac{1}{2}(5)-2=\frac{1}{2} \\ f(7)&=\frac{1}{2}(7)-2=\frac{3}{2} \\ f(9)&=\frac{1}{2}(9)-2=\frac{5}{2}\end{align*}$
::f(3)=12(3)-212f(5)-2=12f(7)=12(7)-2=32f(9)=12(9)-2=52

Step 3: Calculate the area of each rectangle by multiplying each height by the base width .
::第3步:计算每个矩形的面积,将每一高度乘以基宽。

\begin{aligned} 2 \cdot - \frac{1}{2} & = - 1 \\ 2 \cdot \frac{1}{2} & = 1 \\ 2 \cdot \frac{3}{2} & = 3 \\ 2 \cdot \frac{5}{2} & = 5 \end{aligned}

$\begin{align*}2 \cdot -\frac{1}{2} &=-1 \\ 2 \cdot \frac{1}{2} &=1 \\ 2 \cdot \frac{3}{2} &=3 \\ 2 \cdot \frac{5}{2} &=5\end{align*}$

Step 4: Add the four areas together to get the total approximation for the area under the given curve.
::第4步:将四个区域加在一起,以获得给定曲线下区域的总近似值。

\begin{aligned} Area = - 1 + 1 + 3 + 5 = 8 square units \end{aligned}

$\begin{align*}\text{Area} =-1+1+3+5=8 \text{ square units}\end{align*}$
::面积=1+1+3+5=8平方单位

Example 2
::例2

Evaluate the exact area under the curve in Example 1 using the area formula for a triangle.
::使用三角形的区域公式评价例1曲线下的确切区域。

Solution:
::解决方案 :

Remember that the area below the $\begin{align*}x\end{align*}$ -axis is negative, while the area above the $\begin{align*}x\end{align*}$ -axis is positive.
::记住 X 轴下的面积为负, X 轴上方的面积为正。

lesson content

Below the $\begin{align*}x\end{align*}$ - axis : The base of the triangle is 3 units, and the height is -1.5 units.
::x 轴下方: 三角形的底部是 3 个单位, 高度是 - 1. 5 个单位。

\begin{aligned} \frac{1}{2} \cdot 3 \cdot - 1.5 = - \frac{9}{4} \end{aligned}

$\begin{align*}\frac{1}{2} \cdot 3 \cdot -1.5=-\frac{9}{4}\end{align*}$

Above the $\begin{align*}x\end{align*}$ - axis : The base of the triangle is 5 units, and the height is 2.5 units.
::X 轴之上:三角形的底部为 5 个单位,高度为 2.5 个单位。

\begin{aligned} \frac{1}{2} \cdot 5 \cdot 2.5 = \frac{25}{4} \end{aligned}

$\begin{align*}\frac{1}{2} \cdot 5 \cdot 2.5=\frac{25}{4}\end{align*}$

The total area under the curve between 1 and 9 is
::1至9之间曲线下的总面积为

\begin{aligned} - \frac{9}{4} + \frac{25}{4} = \frac{16}{4} = 4 square units . \end{aligned}

$\begin{align*}-\frac{9}{4}+ \frac{25}{4} =\frac{16}{4}=4 \text{ square units}.\end{align*}$
::- 94+254=164=4平方单位。

As expected, the approximated area under the curve is not as accurate the the exact area.
::如预期,曲线下大致的面积没有准确到准确的面积。

Example 3
::例3

Logan travels by bike at 20 mph for 3 hours. Then she gets in a car and drives 60 mph for 2 hours. Sketch both the distance vs. time graph and the rate vs. time graph. Use an area under the curve argument to connect the two graphs.
::洛根乘自行车在20mph旅行3小时。然后她乘车驾驶60mph 2小时。绘制距离图和时间图。使用曲线参数下的一个区域连接两个图形。

Solution:
::解决方案 :

Distance vs. Time:
::距离与时间 :

Rate vs. Time:
::比率与时间 :

The slope of the 1st graph is 20 from 0 to 3, and then 60 from 3 to 5. The 2nd graph is a graph of the slopes from the 1st graph. If you calculate the area of the 2nd graph at the key points 0, 1, 2, 3, 4, and 5, you will see that they align perfectly with the points on the 1st graph.
::第一个图的斜度是 0 到 3 的 20, 然后是 60 从 3 到 5 的 60 。第二个图是第一个图的斜度。如果您在关键点 0, 1, 2, 3, 4 和 5 上计算第二个图的面积, 将会看到它们与第一个图的点完全吻合。

$\begin{align}x\end{align}$	*Area under curve from 0 to* $\begin{align}x\end{align}$
0	0
1	20
2	40
3	60
4	120
5	180

Example 4
::例4

Recall the problem from the Introduction: T he velocity of the car in meters per second can be modeled by a quadratic for the 1st 8 seconds of acceleration $\begin{align*}v(t)=t^2\end{align*}$ .
::回顾导言中的问题:车速每秒以米计,可以用加速度1至8秒的二次方位模型模拟,比(t)=2加速度1至8秒。

Solution:
::解决方案 :

T he area under the curve is equal to the total distance traveled in the 1st 8 seconds. Since the quadratic is a curve, you must choose the number of subintervals you want to use and whether you want right- or left-handed boxes for estimating. Suppose you choose 8 left-handed boxes of width equal to 1.
::曲线下的区域等于在第1 8 秒中穿越的总距离。由于二次曲线是一个曲线,您必须选择您想要使用的次间距数,以及您是否想要用右手或左手框来估算。如果您选择8个左手宽度等于 1 的宽度箱,则您必须选择要使用的次间距数。

lesson content

$\begin{align}x\end{align}$	0	1	2	3	4	5	6	7
Area of box to the right	1⋅0=0	1⋅1=1	1⋅4=4	1⋅9=9	1⋅16=16	1⋅25=25	1⋅36=36	1⋅49=49

The approximate sum is $\begin{align*}1+4+9+16+25+36+49=140\end{align*}$ . This means that the car traveled approximately 140 meters in the 1st 8 seconds.
::大约为1+4+9+16+25+36+49=140,这意味着汽车在1 8秒内行驶约140米。

Example 5
::例5

Approximate the area under the curve using 8 subintervals and right endpoints for the function $\begin{align*}f(x)=3x^2-1, \ -1 \le x \le 7\end{align*}$ .
::使用函数 f( x) = 3x2 - 1, - 1 ×x+7 的 8 次中间值和右端端点接近曲线下的区域。

Solution:
::解决方案 :

While a graph is helpful to visualize the problem, and drawing each box can help give meaning to each summand, this is not always necessary.
::虽然图表有助于直观地看待问题,绘制每个方框有助于使每个摘要具有意义,但这并不总是必要的。

Step 1: Since there are going to be 8 subintervals over the total interval of $\begin{align*}-1 \le x \le 7\end{align*}$ , each interval is going to have a width of 1.
::第1步:由于在-1x+7的总间隔期间将有8个次隔热器,每个间隔的宽度将为1。

\begin{aligned} Δ x = \frac{7 - (- 1)}{8} = \frac{8}{8} = 1 \end{aligned}

$\begin{align*}\Delta x= \frac{7-(-1)}{8}=\frac{8}{8}=1\end{align*}$
::*x=7-(-1)8=88=1

The right-hand endpoint $\begin{align*}x\end{align*}$ -values are 0, 1, 2, 3, 4, 5, 6, and 7.
::右手端点X值为 0, 1, 2, 2, 3, 4, 5, 6, 7。

Step 2: The corresponding function value for these $\begin{align*}x\end{align*}$ -values are:
::第2步:这些X值的相应函数值是:

\begin{aligned} f (0) & = 3 \cdot 0^{2} - 1 = - 1 \\ f (1) & = 3 \cdot 1^{2} - 1 = 2 \\ f (2) & = 3 \cdot 2^{2} - 1 = 11 \\ f (3) & = 3 \cdot 3^{2} - 1 = 26 \\ f (4) & = 3 \cdot 4^{2} - 1 = 47 \\ f (5) & = 3 \cdot 5^{2} - 1 = 74 \\ f (6) & = 3 \cdot 6^{2} - 1 = 107 \\ f (7) & = 3 \cdot 7^{2} - 1 = 146 \end{aligned}

$\begin{align*}f(0) &= 3 \cdot 0^2 -1 =-1\\ f(1) &= 3 \cdot 1^2 -1 =2\\ f(2) &= 3 \cdot 2^2 -1 =11\\ f(3) &= 3 \cdot 3^2 -1 =26\\ f(4) &= 3 \cdot 4^2 -1 =47\\ f(5) &= 3 \cdot 5^2 -1 =74\\ f(6) &= 3 \cdot 6^2 -1 =107\\ f(7) &= 3 \cdot 7^2 -1=146\end{align*}$
::f(0)=302-111f(1)=312-1=2f(2)=322-1=1=11f(3)=332-1=26f(4)=342-1=47f(5)=342-1=1=47f(5)=352-1=74f(6)=362-1=107f(7)=3721=146

Step 3: The area of each rectangle is -1, 2, 11, 26, 47, 74, 107, and 146.
::第3步:每个矩形的面积为-1、2、11、26、47、74、107和146。

Step 4: The area under the curve is
::第4步:曲线下的区域为

\begin{aligned} Area = \sum_{i = 1}^{8} f (c_{i}) \cdot Δ x = - 1 + 2 + 11 + 26 + 47 + 74 + 107 + 146 = 412 square units . \end{aligned}

$\begin{align*}\text{Area}=\sum_{i=1}^{8} f(c_i) \cdot \Delta x= -1+2+11+26 +47 + 74 + 107 +146 = 412 \text{ square units} .\end{align*}$
::面积=18f(ci)x1+2+11+26+47+74+107+146=412平方单位。

Example 6
::例6

Approximate the area under the curve using 8 subintervals and left endpoints for the function: $\begin{align*}f(x)=\frac{4}{x}+3, \ 2 \le x \le 6\end{align*}$ .
::使用 8 个次intervals 和函数的左端点 : f( x) = 4x+3 、 2x = 6 , 接近曲线下的区域。

Solution:
::解决方案 :

Step 1: Since there are going to be 8 subintervals over the total interval of $\begin{align*}2 \le x \le 6\end{align*}$ , each interval is going to have a width of $\begin{align*}\frac{1}{2}\end{align*}$ .
::第1步:由于在总间隔为2x6的总间隔内将有8个次间隔,每个间隔的宽度将为12。

\begin{aligned} Δ x = \frac{6 - 2}{8} = \frac{4}{8} = \frac{1}{2} \end{aligned}

$\begin{align*}\Delta x= \frac{6-2}{8}=\frac{4}{8}=\frac{1}{2}\end{align*}$
::X=6-28=48=12

Since the rectangles hit the curve on the left- hand corner, the 1st intersection point will be at the start of the interval, 2. The left-hand endpoint $\begin{align*}x\end{align*}$ -values are 2, 2.5, 3, 3.5, 4, 4.5, 5, and 5.5.
::2. 左手端点X值为2、2.5、3、3.5、4、4.5、5和5.5。

Step 2: The corresponding function value for these $\begin{align*}x\end{align*}$ -values are:
::第2步:这些X值的相应函数值是:

\begin{aligned} f (2) & = \frac{4}{2} + 3 = 5 \\ f (2.5) & = \frac{4}{2.5} + 3 = \frac{8}{5} \\ f (3) & = \frac{4}{3} + 3 = \frac{13}{3} \\ f (3.5) & = \frac{4}{3.5} + 3 = \frac{29}{7} \\ f (4) & = \frac{4}{4} + 3 = 4 \\ f (4.5) & = \frac{4}{4.5} + 3 = \frac{35}{9} \\ f (5) & = \frac{4}{5} + 3 = \frac{19}{5} \\ f (5.5) & = \frac{4}{5.5} + 3 = \frac{41}{11} \end{aligned}

$\begin{align*}f(2) &= \frac{4}{2} +3 =5\\ f(2.5) &= \frac{4}{2.5} +3 =\frac{8}{5}\\ f(3) &= \frac{4}{3} +3 =\frac{13}{3}\\ f(3.5) &= \frac{4}{3.5} +3 =\frac{29}{7} \\ f(4) &= \frac{4}{4} +3 =4\\ f(4.5) &= \frac{4}{4.5} +3 =\frac{35}{9}\\ f(5) &= \frac{4}{5} +3= \frac{19}{5} \\ f(5.5) &= \frac{4}{5.5} +3=\frac{41}{11} \end{align*}$

::f(2)=42+3=5f(2.5)=42.5+3=3=3=85f(3)=43+3=133f(3.5)=43.5+3=3=297f(4)=44+3=3=4f(4.5)=44.5+3=3=359f=359f=45+3=3=195f(5.5)=45.5+3=4111

Step 3: The area of each rectangle is $\begin{align*}\frac{5}{2},\frac{4}{5},\frac{13}{6},\frac{29}{14},2,\frac{35}{18}, \frac{19}{10}, \text{ and } \frac{41}{22}.\end{align*}$
::第3步:每个矩形的面积为52 45 136 2914 2 3518 1910和4122。

Step 4: The area under the curve is:
::第4步:曲线下的区域为:

\begin{aligned} Area = \sum_{1}^{20} f (c_{i}) \cdot Δ x & = f (1) \cdot 0.1 + f (1.1) \cdot 0.1 + f (1.2) \cdot 0.1 + \dots + f (2.9) \cdot 0.1 \\ = 0.1 (f (1) + f (1.1) + \dots f (2.9)) \\ \approx 12.471 square units . \end{aligned}

$\begin{align*}\text{Area}= \sum_{1}^{20} f(c_i)\cdot \Delta x &=f(1) \cdot 0.1+f(1.1) \cdot 0.1+f(1.2) \cdot 0.1+ \cdots +f(2.9) \cdot 0.1 \\ &=0.1\left(f(1)+f(1.1)+ \cdots f(2.9) \right) \\ & \approx 12.471 \text{ square units}.\end{align*}$
::区域 120f( ci) =x=f(1)_0.1+f(1.1)_0.1+f(1)_0.1+f( 1.2)_0.1)_1*f( 2.9)_0.1=0.1(f)+f( 1.1)_(f)_(2.9)\\12.471平方单位。

Example 7
::例7

Approximate the area under the curve using 20 subintervals and left endpoints for the function $\begin{align*}f(x)=x^x, \ 1 \le x \le 3\end{align*}$ .
::使用 20 个次intervals 和函数 f(x) =xxxx, 1xx% 3 的左端点, 接近曲线下的区域。

Solution:
::解决方案 :

When the number of subintervals gets large and the subintervals get extremely narrow, it will be impossible to draw an accurate picture. This is why using summation notation and thinking through what the indices and the argument will be is incredibly important.
::当次隔热器数量大而次隔热器数量极小时,将无法得出准确的图片。这就是为什么使用总和符号和思考指数和参数将会是何等重要的原因。

Step 1: Since there are going to be 20 subintervals over the total interval of $\begin{align*}1 \le x \le 3\end{align*}$ , each interval is going to have a width of 0.1.
::第1步:由于在总间隔为1x3的间隔内将有20个次间隔,每个间隔的宽度将为0.1。

\begin{aligned} Δ x = \frac{3 - 1}{20} = \frac{2}{20} = 0.1 \end{aligned}

$\begin{align*}\Delta x= \frac{3-1}{20}=\frac{2}{20}=0.1\end{align*}$
::x=3-120=220=0.1

The left-hand endpoint $\begin{align*}x\end{align*}$ -values will consist of 1, 1.1, 1.2, 1.3, 1.4, ... .
::左手端点X值为1、1.1、1.2、1.3、1.4、...。

Step 2: The area under the curve is
::第2步:曲线下的区域为

\begin{aligned} Area = \sum_{1}^{20} f (c_{i}) \cdot Δ x & = f (1) \cdot 0.1 + f (1.1) \cdot 0.1 + f (1.2) \cdot 0.1 + \dots + f (2.9) \cdot 0.1 \\ = 0.1 (f (1) + f (1.1) + \dots f (2.9)) \\ \approx 12.471 square units . \end{aligned}

$\begin{align*}\text{Area}=\sum_{1}^{20} f(c_i) \cdot \Delta x &=f(1) \cdot 0.1+f(1.1) \cdot 0.1+f(1.2) \cdot 0.1+ \cdots +f(2.9) \cdot 0.1 \\ &=0.1(f(1)+f(1.1)+ \cdots f(2.9)) \\ & \approx 12.471 \text{ square units}.\end{align*}$
::区域 120f( ci) =x=f(1)_0.1+f(1.1)_0.1+f(1)_0.1+f( 1.2)_0.1)_1*f( 2.9)_0.1=0.1(f)+f( 1.1)_(f)_(2.9)\\12.471平方单位。

Summary
::摘要

Riemann Sums can be used to approximate the area under a curve.
::Riemann Sums可用于接近曲线下的区域。

::Riemann Sums可用于接近曲线下的区域。

The area under a curve is calculated by $\begin{align*}\text{Area}=\sum_{i=1}^{n}{f(c_i) \cdot \Delta x },\end{align*}$ where $\begin{align*}c_i\end{align*}$ is the $\begin{align*}x\end{align*}$ -value of the point of the box that hits the curve, $\begin{align*}f(c_i)\end{align*}$ is the $\begin{align*}y\end{align*}$ -value at that point, and $\begin{align*}\Delta x\end{align*}$ is the width of the box .
::曲线下的区域由“区域”=%i=1nf(ci)%x计算,其中“c”是击入曲线的框点的x值,“f(ci)”是该点的y值,“x”是该框的宽度。

::曲线下的区域由“区域”=%i=1nf(ci)%x计算,其中“c”是击入曲线的框点的x值,“f(ci)”是该点的y值,“x”是该框的宽度。

For a curve defined on the interval $\begin{align*}[a,b]\end{align*}$ , $\begin{align*}\Delta x = \frac{b-a}{n},\end{align*}$ where $\begin{align*}n\end{align*}$ is the number of rectangles or boxes.
::对于在间距[a,b], x=b-an, 其中 n 是矩形或框数的曲线。

::对于在间距[a,b], x=b-an, 其中 n 是矩形或框数的曲线。

A definite integral is the limit of a sum as the number of summands increases to infinity.
::确定的组成部分是总和的限度,因为总和数增加至无限。

::确定的组成部分是总和的限度,因为总和数增加至无限。

Review
::回顾

1. Approximate the area under the curve using 8 subintervals and right endpoints for the function $\begin{align*}f(x)=x^2-x+1, \ 0 \le x \le 8\end{align*}$ .
::1. 使用函数 f(x) =x2-x+1, 0x+8 的8个次中间值和右端端点,接近曲线下的区域。

2. Approximate the area under the curve using 8 subintervals and left endpoints for the function $\begin{align*}f(x)=x^2-2x+1, \ -4 \le x \le 4\end{align*}$ .
::2. 曲线下区域近似,使用8个次中间值,函数 f(x) =x2-2x+1,- 4xx+4的左端点。

3. Approximate the area under the curve using 20 subintervals and left endpoints for the function $\begin{align*}f(x)=\sqrt{x+3}, \ 0 \le x \le 4\end{align*}$ .
::3. 曲线下区域近距离,使用20个次中间值,函数 f(x) =x+3, 0x+4 的左端点。

4. Approximate the area under the curve using 100 subintervals and left endpoints for the function $\begin{align*}f(x)=\sqrt{x+3}, \ 0 \le x \le 4\end{align*}$ . Compare to your answer from Number 3.
::4. 曲线下的区域近似,使用100个次中间值,函数 f(x) =x+3, 0x+4 的左端点为 f(x) =x+3, 0x=4. 与您从第 3 号中回答的比较。

5. Approximate the area under the curve using 8 subintervals and left endpoints for the function $\begin{align*}f(x)=\cos(x), \ 0 \le x \le 4\end{align*}$ .
::5. 曲线下区域近似,使用8个次中间值,函数 f(x) = cos(x), 0x4 的左端点为 f(x) = cos(x), 0x4 。

6. Approximate the area under the curve using 20 subintervals and left endpoints for the function $\begin{align*}f(x)=\cos(x), \ 0 \le x \le 4\end{align*}$ .
::6. 曲线下区域近似,使用20个次中间值和左端点,用于函数 f(x) = cos(x), 0x4。

7. Approximate the area under the curve using 100 subintervals and left endpoints for the function $\begin{align*}f(x)=\cos(x), \ 0 \le x \le 4\end{align*}$ .
::7. 曲线下区域近似,使用100次间距,函数 f(x) = cos(x), 0x4 的左端点。

The following graph shows the rate (in miles per hour) vs. time (in hours) for a car:
::下图显示汽车的车速(每小时英里)与时间(每小时)之比:

8. Describe what is happening with the car.
::8. 描述汽车的情况。

9. How far did the car travel in 5 hours?
::9. 汽车在5小时内行驶到多远?

The following graph shows the rate (in feet per second) vs. time (in seconds) for a car:
::下图显示汽车的速率(每秒每英尺)相对于时间(每秒)的速率:

10. Describe what is happening with the car. In particular, what is happening in the 1st 3 seconds?
::10. 描述一下汽车的情况,特别是第1秒3秒的情况?

11. How far did the car travel in 5 seconds?
::11. 5秒钟内车程有多远?

The following graph shows the function $\begin{align*}f(x)=-(x-4)^2+16\end{align*}$ , which represents the rate (in feet per second) vs. time (in seconds) for a runner:
::下图显示函数 f(x) (x- 4) 2+16, 该函数代表运行者的速度( 每秒以英尺计) 相对于时间( 秒数) :

12. Describe what is happening with the runner. In particular, what happens after 4 seconds?
::12. 描述赛跑者的情况,特别是4秒后会发生什么情况?

13. Use rectangles to approximate the total distance (in feet) that the runner traveled in the 8 seconds. Try to get as good an approximation as possible.
::13. 使用矩形来估计跑者在8秒内行驶的总距离(以足计),尽量取得良好的近似值。

14. How do integrals relate to sums?
::14. 整体件与总额有何关系?

Review (Answers )
::回顾(答复)

Please see the Appendix.
::请参看附录。

章节大纲

常规

曲线下的区域

Introduction ::导言

Area Under a Curve ::曲线下的区域

Relationship Between an Integral and the Area Under a Curve ::整体关系和曲线下 " 区域 " 之间的关系

Examples ::实例

Summary ::摘要

Review ::回顾

Review (Answers ) ::回顾(答复)

Introduction
::导言

Area Under a Curve
::曲线下的区域

Relationship Between an Integral and the Area Under a Curve
::整体关系和曲线下 " 区域 " 之间的关系

Examples
::实例

Summary
::摘要

Review
::回顾

Review (Answers )
::回顾(答复)