Course: 5.7 缺陷综合体的一些基本属性

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缺乏自定义集成体的某些基本属性

Given what you know about the definite integral as the limit of a , see if you can determine, before you start this lesson, how the definite integral of the sum or difference of two functions could be determined. What other properties of definite integrals can you think of?
::考虑到你对确定整体的极限所知道的,看看在开始这一课之前,你是否可以确定如何确定两个函数的总和或差的确定整体。你能想到确定整体的哪些其他属性?

Properties of Definite Integrals
::限制综合体的属性

Repeated here are a few definitions that are useful when evaluating definite integrals:
::这里重复了几个定义,这些定义在评估确定的整体性时是有用的:

If $\begin{aligned} f (x) \end{aligned}$ is an integrable function on the closed interval $\begin{aligned} [a, b] \end{aligned}$ , then:
::如果 f(x) 是封闭间隔[a,b] 上的不可调试函数,则:

Definition: $\begin{aligned} \int_{a}^{a} f (x) d x = 0 \end{aligned}$ if $\begin{aligned} f (a) \end{aligned}$ exists.
::定义:*aaf(x)dx=0,如果f(a)存在的话。

Definition: If $\begin{aligned} b > a \end{aligned}$ , then $\begin{aligned} \int_{b}^{a} f (x) d x = - \int_{a}^{b} f (x) d x \end{aligned}$ .
::定义:如果 b>a,那么baf(x)dx*abf(x)dx。

The above definitions as well as the following rules that mirror the rules for indefinite integrals, will enable the evaluation of a wide variety of definite integrals.
::上述定义以及下列反映无限期整体体规则的规则,将有助于对各种明确的整体体进行评价。

Evaluating Definite Integrals

::评估缺陷综合症

If $\begin{aligned} f (x) \end{aligned}$ and $\begin{aligned} g (x) \end{aligned}$ are both integrable functions on the closed interval $\begin{aligned} [a, b] \end{aligned}$ , then:
::如果 f(x) 和 g(x) 在封闭间隔[a,b] 上同时具有不可磨灭的函数,则:

$\begin{aligned} \int_{a}^{b} k \cdot f (x) d x = k \int_{a}^{b} f (x) d x \end{aligned}$ , for any real number $\begin{aligned} k \end{aligned}$ .
::{abk}f(x)dx=kabf(x)dx,任何实际数字 k。

$\begin{aligned} \int_{a}^{b} [f (x) \pm g (x)] d x = \int_{a}^{b} f (x) d x \pm \int_{a}^{b} g (x) d x \end{aligned}$ .
::-=YTET -伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=- 翻译:

$\begin{aligned} \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x \end{aligned}$ , for $\begin{aligned} a < b < c \end{aligned}$ .
::*acf(x)dx*abf(x)dx*bcf(x)dxx(x)dxx,用于 <b <c > 。

Let's use the properties above to help you compute $\begin{aligned} \int_{1}^{4} (x - \sqrt{x}) d x \end{aligned}$ .
::让我们使用上面的属性来帮助您计算 #% 14( x- x) dx 。

The rule for the difference of two functions can be used to evaluate this definite integral.
::两种职能的区别规则可用来评价这一明确的整体性。

\begin{aligned} \int_{1}^{4} (x - \sqrt{x}) d x & = \int_{1}^{4} x d x - \int_{1}^{4} \sqrt{x} d x \\ = \frac{x^{2}}{2}]_{1}^{4} - \frac{2}{3} x^{\frac{3}{2}}]_{1}^{4} \\ = (8 - \frac{1}{2}) - \frac{2}{3} (8 - 1) \\ = \frac{15}{2} - \frac{14}{3} \\ \int_{1}^{4} (x - \sqrt{x}) d x & = \frac{17}{6} \end{aligned}

::14(x-x)dx 14xdx14xxx=x22)14-23x32)14=(8-12)-23(8-1)=152-143}14(x-x)dx=176

Examples
::实例

Example 1
::例1

Earlier, you were asked to think about how to find the integral of a sum or difference given what you know about Riemann sums. If the limit of Riemann sums were used to compute the integral of the sum or difference of two functions, the result could be determined by using a separate sum for each function. This is what the properties above have established. But we now have an easier method, using antiderivatives, to evaluate definite integrals. The properties of definite integrals are very similar to the properties of sums and limits.
::早些时候, 您被要求思考如何找到一个总和或差数的有机体, 取决于您对 Riemann 数额的了解。如果使用 Riemann 数额的限度来计算两个函数的总和或差数的有机体, 则结果可以通过对每个函数使用一个单独的总和来确定。这是上面的属性所建立的。但是我们现在有一个比较容易的方法, 使用抗降解法来评估确定的综合体。确定的综合体的特性与数值和限值的特性非常相似。

Example 2
::例2

Compute $\begin{aligned} \int_{0}^{\frac{π}{2}} (x + \cos x) d x \end{aligned}$ .
::计算 02(x+cosx)dx。

The rule for the sum of two functions can be used to evaluate this definite integral.
::两项职能之和的规则可用来评价这一明确的整体性。

\begin{aligned} \int_{0}^{\frac{π}{2}} (x + \cos x) d x & = \int_{0}^{\frac{π}{2}} (x) d x + \int_{0}^{\frac{π}{2}} (\cos x) d x \\ = \frac{x^{2}}{2}]_{0}^{\frac{π}{2}} + \frac{\sin x}{1}]_{0}^{\frac{π}{2}} \\ = \frac{π^{2}}{8} + 1 \\ \int_{0}^{\frac{π}{2}} (x + \cos x) d x & = \frac{π^{2} + 8}{8} \end{aligned}

::========================x22==================================================================================================================================xxxxxxx=x+=x+====================================================================================================================================================================================================================================================================================================================================================

Example 3
::例3

Show that $\begin{aligned} \int_{2}^{5} (x^{3} + 4) d x = \int_{2}^{4} (x^{3} + 4) d x + \int_{4}^{5} (x^{3} + 4) d x \end{aligned}$ .
::显示 @ 25( x3+4) dx @ 24( x3+4) dx @ 45( x3+4) dx。

To show that the equality holds, just evaluate separately the two components on the right hand side of the equation, and then add and compare with the left hand side:
::为了证明平等有效,只需分别评估方程式右侧的两个组成部分,然后与左手侧添加和比较:

	I	II	Comment
$\begin{aligned} \int_{2}^{5} (x^{3} + 4) d x \end{aligned}$ ::25(x3+4)dx	$\begin{aligned} \int_{2}^{4} (x^{3} + 4) d x \end{aligned}$ ::24(x3+4)dx	$\begin{aligned} \int_{4}^{5} (x^{3} + 4) d x \end{aligned}$ ::45(x3+4)dx	. . . Each function is integrable ::. . . 每个函数都是不可磨灭的
$\begin{aligned} \int_{2}^{5} x^{3} d x + \int_{2}^{5} 4 d x \end{aligned}$ ::25x3dx 254dx	$\begin{aligned} \int_{2}^{4} x^{3} d x + \int_{2}^{4} 4 d x \end{aligned}$ ::24x3dx 2444dx	$\begin{aligned} \int_{4}^{5} x^{3} d x + \int_{4}^{5} 4 d x \end{aligned}$ ::45x3dx 454dx	. . . Addition property
$\begin{aligned} \frac{x^{4}}{4} \|_{2}^{5} + 4 x \|_{2}^{5} \end{aligned}$ ::x4425+4x25	$\begin{aligned} \frac{x^{4}}{4} \|_{2}^{4} + 4 x \|_{2}^{4} \end{aligned}$ ::x4424+4x24	$\begin{aligned} \frac{x^{4}}{4} \|_{4}^{5} + 4 x \|_{4}^{5} \end{aligned}$ ::x4445+4x45
$\begin{aligned} (\frac{625}{4} - \frac{16}{4}) + (20 - 8) \end{aligned}$	$\begin{aligned} (\frac{256}{4} - \frac{16}{4}) + (16 - 8) \end{aligned}$	$\begin{aligned} (\frac{625}{4} - \frac{256}{4}) + (20 - 16) \end{aligned}$
$\begin{aligned} \frac{609}{4} + 12 \end{aligned}$	$\begin{aligned} \frac{240}{4} + 8 \end{aligned}$	$\begin{aligned} \frac{369}{4} + 4 \end{aligned}$
$\begin{aligned} \frac{657}{4} = \end{aligned}$	$\begin{aligned} \frac{272}{4} + \end{aligned}$	$\begin{aligned} \frac{385}{4} \end{aligned}$
$\begin{aligned} \frac{657}{4} \end{aligned}$	$\begin{aligned} \frac{657}{4} \end{aligned}$

Example 4
::例4

Compute $\begin{aligned} \int_{0}^{15} g (x) d x \end{aligned}$ , where $\begin{aligned} g (x) = {\begin{cases} 5^{x}, 0 \leq x \leq 3 \\ 125, 3 \leq x \leq 5 \\ 1.25 (x - 15)^{2}, 5 \leq x \leq 15 \end{cases} \end{aligned}$ .
::计算 =015g(x)dx, g(x) 5, 0x 3125, 3x51.25(x-15)2, 5x 15。

The function $\begin{aligned} g (x) \end{aligned}$ is a piece-wise continuous function over [0, 15] that requires integration over 3 subintervals, as follows:
::g(x) 函数是在 [ 0, 15] 上方的字符串连续函数,需要在 3 次对数上进行整合,具体如下:

**Subinterval**
	$\begin{aligned} 0 \leq x \leq 3 \end{aligned}$ ::0x% 3	$\begin{aligned} 3 \leq x \leq 5 \end{aligned}$ ::3x5	$\begin{aligned} 5 \leq x \leq 15 \end{aligned}$ ::5x%15
$\begin{aligned} \int_{0}^{15} g (x) d x \end{aligned}$ ::015g( x) dx	$\begin{aligned} = \int_{0}^{3} 5^{x} d x \end{aligned}$ ::035xxxx	$\begin{aligned} + \int_{3}^{5} 125 d x \end{aligned}$ ::35125dx	$\begin{aligned} + \int_{5}^{15} 1.25 (x - 15)^{2} d x \end{aligned}$ ::515.25(x-155)2dx
	$\begin{aligned} = \frac{5^{x}}{\ln 5} \|_{0}^{3} \end{aligned}$ ::=5xln503	$\begin{aligned} + 125 x \|_{3}^{5} \end{aligned}$ ::+125x35	$\begin{aligned} + \frac{1.25}{3} (x - 15)^{3} \|_{5}^{15} \end{aligned}$ ::+1.253(x--155)3515
	$\begin{aligned} = \frac{125}{\ln 5} - \frac{1}{\ln 5} \end{aligned}$ ::=125ln__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________	$\begin{aligned} + 625 - 375 \end{aligned}$	$\begin{aligned} + 0 + \frac{1250}{3} \end{aligned}$
	$\begin{aligned} = \frac{124}{\ln 5} \end{aligned}$ ::=124ln5	$\begin{aligned} + 250 \end{aligned}$	$\begin{aligned} + \frac{1250}{3} \end{aligned}$
	$\begin{aligned} = 77.0 \end{aligned}$	$\begin{aligned} + 250 \end{aligned}$	$\begin{aligned} + 416.7 \end{aligned}$
$\begin{aligned} \int_{0}^{15} g (x) d x \end{aligned}$ ::015g( x) dx	$\begin{aligned} = 743.7 \approx 744 \end{aligned}$

Review
::回顾

For #1-12, use antiderivatives to compute the definite integral.
::对于 # 1-12, 使用抗降解剂来计算确定的整体。

$\begin{aligned} \int_{4}^{9} (\frac{3}{\sqrt{x}}) d x \end{aligned}$
::49(3x)dx

$\begin{aligned} \int_{0}^{1} (t - t^{2}) d t \end{aligned}$
::01(t- t2)dt

$\begin{aligned} \int_{2}^{5} (\frac{1}{\sqrt{x}} + \frac{1}{\sqrt{2}}) d x \end{aligned}$
::================================================================================================================================================== =================================================================================================================================================================================================================================

$\begin{aligned} \int_{0}^{1} 4 (x^{2} - 1) (x^{2} + 1) d x \end{aligned}$
::014(x2- 1)(x2+1)dx

$\begin{aligned} \int_{2}^{8} (\frac{4}{x} + x^{2} + x) d x \end{aligned}$
::28(4x+x2+x)dx

$\begin{aligned} \int_{2}^{4} (e^{3 x}) d x \end{aligned}$
::24(e3x)dx

$\begin{aligned} \int_{1}^{4} \frac{2}{x + 3} d x \end{aligned}$
::142x+3dx

$\begin{aligned} \int_{0}^{\frac{π}{2}} (3 \cos x - 5) d x \end{aligned}$
::02(2) 3cosx-5dx

$\begin{aligned} \int_{0}^{1} (5 x^{5} - 7 x^{2} + 4) d x \end{aligned}$
::01( 5x5- 7x2+4) dx

$\begin{aligned} \int_{0}^{1} 5 x e^{- 2 x^{2}} d x \end{aligned}$
::=015xe-2x2dx

$\begin{aligned} \int_{0}^{\frac{π}{4}} (\sec^{2} x (2 \tan x - 1)) d x \end{aligned}$
::04( sec2x( 2tanx- 1)) dx

$\begin{aligned} \int_{0}^{1} (5 x + 1)^{3} d x \end{aligned}$
::@ 01( 5x+1) 3dx

Find the average value of $\begin{aligned} f (x) = \sqrt{x} \end{aligned}$ over [1, 9].
::查找 [1, 9] 以上 f(x) =x 的平均值。

If $\begin{aligned} f \end{aligned}$ is continuous and $\begin{aligned} \int_{1}^{4} f (x) d x = 9 \end{aligned}$ , show that $\begin{aligned} f \end{aligned}$ takes on the value 3 at least once on the interval [1, 4].
::如果 f 是连续的, 且 14f(x)dx=9, 则显示 f 在间隔 [1, 4] 上至少拿取3 值一次。

Your friend states that there is no area under the curve of $\begin{aligned} f (x) = \sin x \end{aligned}$ on $\begin{aligned} [0, 2 π] \end{aligned}$ since he computed $\begin{aligned} \int_{0}^{2 π} \sin x d x = 0 \end{aligned}$ . Is he correct? Explain your answer.
::您的朋友表示, f( x) =sinx 曲线[ 0, 2] 上没有区域, 因为他计算 = 02sinxdx=0。他是否正确? 请解释您的答复。

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

Section outline

General

缺乏自定义集成体的某些基本属性

Properties of Definite Integrals ::限制综合体的属性

Evaluating Definite Integrals ::评估缺陷综合症

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Review ::回顾

Review (Answers) ::回顾(答复)