Section: 明确整合:变量变化 | 5.9 明确整合:变数变化

Section outline

In computing an indefinite integral, the technique of performing a change of variable, or variable substitution ( $\begin{aligned} u \end{aligned}$ -substitution), is often a way to make a “difficult” integral $\begin{aligned} \int f (x) d x \end{aligned}$ easier to integral $\begin{aligned} \int f (u) d u \end{aligned}$ evaluate. Once the easier integration is performed to obtain an , the “new” variable is replaced in the antiderivative by its equivalent “old” variable expression. But, if the integral being evaluated is a definite integral $\begin{aligned} \int_{a}^{b} f (x) d x \end{aligned}$ , and we want to write the easier definite integral $\begin{aligned} \int_{?}^{?} f (u) d u \end{aligned}$ , do the limits of integration change?
::在计算一个无限期的整体时,进行变数或变数替代(替代)的技术往往是一种使“困难”整体化的方法,更容易进行整体化评价。一旦进行较容易的整合以获得一个,“新”变量在抗衍生变数中被与其等同的“旧”变数表达法所取代。但是,如果所评价的组合是明确的有机化的 {abf(x)dx,而我们想写一个较容易的确定整体化的???f(u)du,那么整合的极限是多少?

Integration by Substitution
::以替代方式的融合

The technique of $\begin{aligned} u \end{aligned}$ -substitution (or change of variable) can also be used when evaluating definite integrals, but it requires the additional step of changing the limits of integration to be consistent. The required formulation is summarized as follows:
::替代技术(或变数的改变)也可以用于评价确定的整体体,但它要求改变整合限度这一额外步骤保持一致。

Problem: Evaluate $\begin{aligned} \int_{a}^{b} f (x) d x \end{aligned}$
::问题:评价 abf(x)dx

Substitution: Let $\begin{aligned} u = g (x) \end{aligned}$ , so that $\begin{aligned} d u = g^{'} (x) d x \end{aligned}$ and $\begin{aligned} f (x) d x = f (g (x)) g^{'} (x) d x \end{aligned}$
::替代 : Let u=g(x), 所以 du=g=xxdx 和 f( x) dx=f( g(x)) g*(x) dx

Transform Problem : $\begin{aligned} \int_{a}^{b} f (x) d x = \int_{a}^{b} f (g (x)) g^{'} (x) d x = \int_{g (a)}^{g (b)} f (u) d u \end{aligned}$
::变换问题 : @ abf( x) dx* *abf( g( x) g( x) dx*g( a) g( b) f( u) du

The goal is to make $\begin{aligned} \int_{g (a)}^{g (b)} f (u) d u \end{aligned}$ easier to evaluate than $\begin{aligned} \int_{a}^{b} f (x) d x \end{aligned}$ .
::目标是使g(a)g(b)f(u)du比abf(x)dx更容易评估。

Take the definite integral $\begin{aligned} \int_{0}^{3} \frac{d x}{4 x + 5} \end{aligned}$ .
::使用确定的整体 03dx4x+5 。

Try to substitute $\begin{aligned} u = 4 x + 5 \end{aligned}$ . Then $\begin{aligned} d u = 4 d x \end{aligned}$ .
::尝试替换 u=4x+5 。然后 du=4dx 。

Lower limit: For $\begin{aligned} x = 0, u = 4 \cdot 0 + 5 = 5 \end{aligned}$
::下限: x=0, u=40+5=5

Upper limit: For $\begin{aligned} x = 3, u = 4 \cdot 3 + 5 = 17 \end{aligned}$ .
::上限: x=3, u=43+5=17。

Therefore
::因此,

$\begin{aligned} \int_{0}^{3} \frac{d x}{4 x + 5} & \int_{5}^{17} \frac{d u}{4 u} \\ = \frac{1}{4} [\ln u]_{5}^{17} \\ \int_{0}^{3} \frac{d x}{4 x + 5} & \int_{5}^{17} \frac{d u}{4 u} \\ = \frac{1}{4} [\ln 17 - \ln 5] \end{aligned}$

::============================================================================================================================================================== ==============================================================================================================================================================================================================================================================================================================================================================

Let’s try the substitution method of definite integrals with a trigonometric integrand .
::让我们试一下用三角分数成份替代确定整体体的方法。

Evaluate $\begin{aligned} \int_{0}^{\frac{π}{4}} \sin 4 x d x \end{aligned}$ .
::评估04sin4xdx。

Let $\begin{aligned} u = 4 x \end{aligned}$ . Then $\begin{aligned} d u = 4 d x \end{aligned}$ .
::Letu=4x,然后du=4dx。

Lower limit: For $\begin{aligned} x = 0, u = 4 \cdot 0 = 0 \end{aligned}$
::下限: x=0, u=40=0

Upper limit: For $\begin{aligned} x = \frac{π}{4}, u = 4 \frac{π}{4} = π \end{aligned}$ .
::上限: x4, u=444。

Therefore
::因此,

$\begin{aligned} \int_{0}^{\frac{π}{4}} \sin 4 x d x & \int_{0}^{π} \frac{\sin u}{4} d u \\ = \frac{1}{4} [- \cos u]_{0}^{π} \\ = \frac{1}{4} [- (- 1) - (- 1)] \\ \int_{0}^{\frac{π}{4}} \sin 4 x d x & = \frac{1}{2} \end{aligned}$

::~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Examples
::实例

Example 1
::例1

Earlier, you were asked what the limits of integration for the new integral are after $\begin{aligned} u \end{aligned}$ -substitution. Are they different from the old limits?
::早些时候,有人问您,在替代后,新整体体的融合限制是什么?它们与旧的限制有区别吗?

They can be! If the $\begin{aligned} u \end{aligned}$ -substitution is $\begin{aligned} u = g (x) \end{aligned}$ , then
::可以!如果u替代是u=g(x),那么

$\begin{aligned} \int_{a}^{b} f (x) d x = \int_{u = g (a)}^{u = g (b)} f (u) d u . \end{aligned}$

::@abf(x)dxu=g(a)u=g(b)f(u)du。

Example 2
::例2

Evaluate $\begin{aligned} \int_{1}^{3} \frac{x}{\sqrt{2 x - 1}} d x \end{aligned}$
::评估 13x2x-1-1dx

Let $\begin{aligned} u = 2 x - 1 \end{aligned}$ . Then $\begin{aligned} d u = 2 d x \end{aligned}$ , or $\begin{aligned} d x = \frac{d u}{2} \end{aligned}$ .
::Letu=2x-1. 然后 du=2dx, 或 dx=du2。

Before we substitute, we need to determine the new limits of integration in terms of the $\begin{aligned} u \end{aligned}$ variable. To do so, we simply substitute the current limits of integration into $\begin{aligned} u = 2 x - 1 \end{aligned}$ :
::在替代之前,我们需要确定u变数的整合新限制。为了这样做,我们只需将目前的整合限制替换为 u=2x-1:

Lower limit: For $\begin{aligned} x = 1, u = 2 (1) - 1 = 1 \end{aligned}$ .
::下限: x=1, u=2(1)- 1=1, 对于 x=1, u=2(1)- 1=1, 下限:

Upper limit: For $\begin{aligned} x = 3, u = 2 (3) - 1 = 5 \end{aligned}$ .
::上限: x=3, u=2(3)-1=5, 上限:

We now substitute $\begin{aligned} u \end{aligned}$ and the associated limits into the integral:
::我们现在将u和相关的限度改成:

$\begin{aligned} \int_{1}^{5} \frac{x}{\sqrt{u}} \frac{d u}{2} \end{aligned}$
::15xudu2

As you can see, the variable $\begin{aligned} x \end{aligned}$ still appears in the integrand. To write it in terms of $\begin{aligned} u \end{aligned}$ , we use the substitution $\begin{aligned} u = 2 x - 1 \end{aligned}$ and solve for $\begin{aligned} x \end{aligned}$ to get, $\begin{aligned} x = \frac{(u + 1)}{2} \end{aligned}$ .
::如您所看到的, 变量 x 仍然出现在整数中。要用 u 写的话, 我们使用 u= 2x- 1 替换, 并解决 x 获取, x= (u+1) 2 。

Substituting back into the integral,
::重置为整体体,

$\begin{aligned} \int_{1}^{3} \frac{x}{\sqrt{2 x - 1}} d x & = \int_{1}^{5} \frac{u + 1}{2 \sqrt{u}} \frac{d u}{2} \\ = \frac{1}{4} \int_{1}^{5} \frac{u + 1}{\sqrt{u}} d u \\ = \frac{1}{4} \int_{1}^{5} (u^{\frac{1}{2}} + u^{- \frac{1}{2}}) d u \\ = \frac{1}{4} {[\frac{2}{3} u^{\frac{3}{2}} + 2 u^{\frac{1}{2}}]}_{1}^{5} \end{aligned}$

::13x2x-1dx15u+12udu2=14}15u+1udu=1415(u12+u-12) du=14[23u32+2u12]15

Applying the by inserting the limits of integration and calculating and simplifying, we get
::通过插入整合、计算和简化的界限,我们得到

$\begin{aligned} \int_{1}^{3} \frac{x}{\sqrt{2 x - 1}} d x & = \frac{1}{4} [(\frac{2}{3} 5^{\frac{3}{2}} + 2 \cdot 5^{\frac{1}{2}}) - (\frac{2}{3} 1^{\frac{3}{2}} + 2 \cdot 1^{\frac{1}{2}})] \\ = \frac{4 \sqrt{5} - 2}{3} \end{aligned}$

::13x2x-1dx=14[(23532+2(512)-(23132+2(1112)]=45-23

We could have chosen the substitution $\begin{aligned} u = \sqrt{2 x - 1} \end{aligned}$ instead. This is left as an exercise in the Review section. Determine for yourself which is the easier approach.
::我们本可以选择 u=2x-1 替代。这将留待审查部分作为练习。请自己决定哪一种方法比较容易。

Example 3
::例3

Evaluate $\begin{aligned} \int_{0}^{\frac{π}{4}} \tan x \sec^{2} x d x \end{aligned}$ .
::评估 04tanxsec2xdx。

Try $\begin{aligned} u = \tan x \end{aligned}$ . Then $\begin{aligned} d u = \sec^{2} x d x \end{aligned}$ .
::试试看,然后是sec2xdx

Lower limit: For $\begin{aligned} x = 0, u = \tan 0 = 0 \end{aligned}$
::下限: x=0, u=tan=0=0

Upper limit: For $\begin{aligned} x = \frac{π}{4}, u = \tan \frac{π}{4} = 1 \end{aligned}$ .
::上限: 对于 x4, u=tan}4=1, 上限: 对于 x4, u=tan=4=1, 上限:

Thus,
::因此,

$\begin{aligned} \int_{0}^{\frac{π}{4}} \tan x \sec^{2} x d x & \int_{0}^{1} u d u \\ = {[\frac{u^{2}}{2}]}_{0}^{1} \\ = \frac{1}{2} \end{aligned}$

::04tanxsec2xdx101udu=[u22]01=12

Review
::回顾

Evaluate the following definite integrals.
::评估以下确定的组成部分。

$\begin{aligned} \int_{1}^{3} \frac{x}{\sqrt{2 x - 1}} d x \end{aligned}$ using the substitution $\begin{aligned} u = \sqrt{2 x - 1} \end{aligned}$ .
::*=13x2x-1dx,使用替换u=2x-1。

$\begin{aligned} \int_{0}^{2} x e^{x^{2}} d x \end{aligned}$ .
::02.xex2dx.

$\begin{aligned} \int_{0}^{\sqrt{π}} x \sin x^{2} d x \end{aligned}$ .
::0xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxdxxxxxxxxxxxxxxxxxxxxxdxxxxxxxx2dxxxxxxxxdxxxxxxxxxxxxdxxxxxxxxxxxxxdxxxxxdxxxxxxxxxxxxdxdxxdxxxxxxxdxxxxxxxxxxxdxxxxxxdxxxxxxdxxxxxxxxxdxdxxxxxxdxxxxxxxxxxdxxxxxxdxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\begin{aligned} \int_{0}^{1} x (x + 5)^{4} d x \end{aligned}$ .
::01x(x+5)4dx。

$\begin{aligned} \int_{0}^{5} 2 x \cos (x^{2}) d x \end{aligned}$ .
::052xcos(x2xxdx)x

$\begin{aligned} \int_{2}^{3} \cos (x) \sin (\sin (x)) d x \end{aligned}$ .
:xxxxxxxxxxxxxxxxxxxxxx)

$\begin{aligned} \int_{π}^{5 π} \sqrt{2 x + 1} d x \end{aligned}$ .
::52x+1dx。

$\begin{aligned} \int_{8}^{10} \frac{\cos (\sqrt{x})}{\sqrt{x}} d x \end{aligned}$ .
::810cos(x)xdxx。

$\begin{aligned} \int_{0}^{\frac{π}{2}} 4 \sin (\frac{x}{3}) d x \end{aligned}$ .
::024sin(x3x)dx。

$\begin{aligned} \int_{0}^{3} x^{2} e^{- x^{3}} d x \end{aligned}$ .
::03x2e -x3dx.

$\begin{aligned} \int_{0}^{1} (3 x - 1)^{27} d x \end{aligned}$ .
::01(3x-1) 27dx

$\begin{aligned} \int_{0}^{1} (5 x^{5} + x^{2})^{3} (25 x^{4} + 2 x) d x \end{aligned}$ .
::01(5x5+x2)3(25x4+2x)dx。

$\begin{aligned} \int_{\frac{π}{2}}^{π} \cos (\frac{x}{2} + π) d x \end{aligned}$ .
::================================================ ============================================================================================================================================================================================================================================================================================================================================================================================================================================================================

$\begin{aligned} \int_{1}^{3} \frac{1}{x^{2}} \sqrt{1 + \frac{1}{x}} d x \end{aligned}$ .
::131x21+1xdxx

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

Integration by Substitution ::以替代方式的融合

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Review ::回顾

Review (Answers) ::回顾(答复)

Integration by Substitution
::以替代方式的融合

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Review
::回顾

Review (Answers)
::回顾(答复)