Section: 指数函数和对数函数的元件 | 7.4 指数函数和对数函数综合体

Section outline

F ormulas for generalized of exponential and make finding the antiderivatives straightforward. How would you determine the volume of an object whose shape is generated by rotating the function $\begin{aligned} f (x) = 5 e^{- 0.5 x} (0 \leq x \leq 3) \end{aligned}$ about the $\begin{aligned} x \end{aligned}$ -axis? Can you find the volume?
::直截了当地找到抗降解剂。您将如何确定函数 f( x) = 5 = -0.5x( 0x) 生成的物体的形状的大小 ? 您能够找到它的大小吗 ?

Integrating Exponential and Logarithmic Functions
::集成指数函数和对数函数

Integrals Involving Exponential Functions
::涉及生态功能的综合系统

For exponential functions, the following derivative-integral relationship holds:
::就指数函数而言,下列衍生物与整体关系:

Given: $\begin{aligned} \frac{d}{d x} [b^{u}] = (\ln b \cdot b^{u}) \frac{d u}{d x} \end{aligned}$
::给定值: ddx[bu] =( lnbbu) dudx

Then: $\begin{aligned} \int b^{u} d u = \frac{1}{\ln b} b^{u} + C \end{aligned}$
::然后:budu=1lnbubu+C

Note: $\begin{aligned} For b = e : \frac{d}{d x} [e^{u}] = e^{u} \frac{d u}{d x} \Leftrightarrow \int e^{u} d u = e^{u} + C \end{aligned}$
::注:b=e:ddx[eu]=eudxeudu=eu+C

The above result is derived from these steps:
::以上是这些步骤的结果:

$\begin{aligned} \frac{d}{d x} [b^{u}] & = (\ln b \cdot b^{u}) \frac{d u}{d x} \\ \int d [b^{u}] & = \int (\ln b \cdot b^{u}) d u \\ b^{u} + C & = \ln b \int b^{u} d u \\ \int b^{u} d u & = \frac{1}{\ln b} b^{u} + C \\ \frac{1}{\ln b} b^{u} + C & = \int b^{u} d u \end{aligned}$

::ddx[bu] = (lnbbu) dudx*d[bu] (lnbbu) bubu+C=lnbbudu*budu=1lnbbu+C1lnbbu+Cbudu

So that, $\begin{aligned} \int b^{u} d u = \frac{1}{\ln b} b^{u} + C \end{aligned}$
::因此,{budu=1ln{bbu+C}

The following problems illustrate how to apply the formula for integrating an exponential function.
::以下问题说明了如何应用公式整合指数函数。

Evaluate: $\begin{aligned} \int 5^{(x - 2.5)} d x \end{aligned}$
::评价:%5(x-2.5)dx

For $\begin{aligned} \int 5^{(x - 2.5)} d x \end{aligned}$ , use $\begin{aligned} u \end{aligned}$ -substitution, and let $\begin{aligned} u = x - 2.5 \end{aligned}$ . Then $\begin{aligned} d u = d x \end{aligned}$ and
::5 (x- 2. 5) dx, 使用 u 替代, let u=x- 2. 5。然后 du= dx and

$\begin{aligned} \int 5^{(x - 2.5)} d x & = \int 5^{u} d u \\ = \frac{1}{\ln 5} 5^{u} + C \\ = \frac{1}{\ln 5} 5^{(x - 2.5)} + C \end{aligned}$

::=5(x-2.5)dx=5udu=1ln5au+C=1ln55(x-2.5)+C

Evaluate: $\begin{aligned} \int_{0}^{2} x^{2} e^{x^{3}} d x \end{aligned}$
::评价: 02x2ex3dx

For $\begin{aligned} \int_{0}^{2} x^{2} e^{x^{3}} d x \end{aligned}$ , use $\begin{aligned} u \end{aligned}$ -substitution, and let $\begin{aligned} u = x^{3} \end{aligned}$ . Then $\begin{aligned} d u = 3 x^{2} d x \end{aligned}$ and
::= 02x2ex3dx,使用 u 替代, let u=x3. 然后 du= 3x2dx和

$\begin{aligned} \int_{0}^{2} x^{2} e^{x^{3}} d x & = \int_{0}^{2^{3}} e^{u} \frac{1}{3} d u \\ = \frac{1}{3} e^{u} |_{0}^{8} \\ = \frac{1}{3} (e^{8} - 1) \end{aligned}$

::02x2ex3dx023eu13du=13eu08=13(e8-1)

Integrals Involving Logarithmic Functions
::包含对对数函数的元件

For logarithmic functions, the following derivative-integral relationship holds:
::对于对数函数,下列衍生物-整体关系存在:

Given: $\begin{aligned} \frac{d}{d x} [\log_{b} u] = \frac{1}{u \ln b} \frac{d u}{d x} \end{aligned}$
::给定值: ddx [logbu] = 1uln bdudx

Then: $\begin{aligned} \int \frac{1}{u (x)} d u = \ln b \cdot \log_{b} | u (x) | + C \end{aligned}$
::然后:1u(x)du=lnblogbu(x)C

Note: $\begin{aligned} For b = e : \frac{d}{d x} [\ln u] = \frac{1}{u} \frac{d u}{d x} \Leftrightarrow \int \frac{1}{u (x)} d u = \ln | u (x) | + C \end{aligned}$
::注: 对于 b=e:ddx[ lnu] = 1udududx1u(x)du=lnu(x) C

The above result is derived from these steps:
::以上是这些步骤的结果:

$\begin{aligned} \frac{d}{d x} [\log_{b} u] & = \frac{1}{u \ln b} \frac{d u}{d x} \\ \int d [\log_{b} u] & = \int \frac{d u}{u \ln b} \\ \log_{b} u + C & = \frac{1}{\ln b} \int \frac{d u}{u} \\ \ln b \cdot \log_{b} u + C & = \int \frac{d u}{u} \end{aligned}$

::ddx [logbu] = 1uln bdudxd [logbu] duuln logbu+C= 1ln bduulnbb logbu+Cduu

So that $\begin{aligned} \int \frac{1}{u (x)} d u = \ln b \cdot \log_{b} | u (x) | + C \end{aligned}$ .
::因此,1u(x)du=lnblogbu(x)C。

When the base is $\begin{aligned} b = e \end{aligned}$ , the logarithm is the natural logarithm, with the result
::当 b==e 时,对数为自然对数,结果为

$\begin{aligned} \int \frac{1}{u (x)} d u = \ln | u (x) | + C . \end{aligned}$

::1u(x)du=lnu(x)C。

Notice that the logarithm function is the of the integrand of the type $\begin{aligned} \frac{d u}{u} \end{aligned}$ . This means that it is important to recognize this form, especially when dealing with a rational integrand.
::注意对数函数是 duu 类型的正数。这意味着必须确认该形式, 特别是在处理理性正数时。

Examples
::实例

Example 1
::例1

Earlier, you were asked about the volume of an object whose shape is generated by rotating the function $\begin{aligned} f (x) = 5 e^{- 0.5 x} (0 \leq x \leq 3) \end{aligned}$ about the $\begin{aligned} x \end{aligned}$ -axis.
::早些时候,有人问及一个物体的体积,该物体的形状是通过旋转函数 f(x)=5e-0.5x(0xx)生成的。

Use the derivative results of the last concept: $\begin{aligned} \frac{d}{d x} [e^{u}] = e^{u} \frac{d u}{d x} \Leftrightarrow \int e^{u} d u = e^{u} + C \end{aligned}$ .
::使用最后一个概念的衍生结果:ddx[eu]=eudxeudu=eu+C。

The volume is given by:
$\begin{aligned} V & = \int_{0}^{3} π f^{2} (x) d x = \int_{0}^{3} 25 π (e^{- 0.5 x})^{2} d x = 25 π \int_{0}^{3} e^{- x} d x = - 25 π e^{- x} |_{0}^{3} \\ V & = 25 π (1 - e^{- 3}) \approx 75 {in}^{3} \end{aligned}$

::给定的音量为: V03f2(x)dx0325( e-0.5x) 2dx=2503e- xdx255e- x03V=25( 1- e-3) @75英寸3

Example 2
::例2

Evaluate: $\begin{aligned} \int \frac{1}{x + 1} d x \end{aligned}$
::评价:% 1x+1dx

In general, whenever there is an integral that has a rational function as an integrand, it might be possible that it can be integrated with the result being a natural logarithm.
::一般而言,只要有一个整体体具有作为一元的合理功能,它就有可能与自然对数相结合。

First look at using $\begin{aligned} u \end{aligned}$ -substitution to see if the integrand can be transformed.
::首先看看用u替代来查看是否能改变整数。

Let $\begin{aligned} u = x + 1 \end{aligned}$ , then $\begin{aligned} d u = d x \end{aligned}$ . Then
::Letu=x+1, 然后du=dx。然后

$\begin{aligned} \int \frac{1}{x + 1} d x & = \int \frac{d u}{u} \\ = \ln | u | + C \\ = \ln | x + 1 | + C \end{aligned}$

::1x+1dxdxduu=lnuC=lnx+1C

Remark: The integral must use the absolute value symbol because although $\begin{aligned} x \end{aligned}$ may have negative values, the domain of $\begin{aligned} \ln (x + 1) \end{aligned}$ is restricted to $\begin{aligned} x \geq 0 \end{aligned}$ .
::备注:整体体必须使用绝对值符号,因为虽然x可能有负值,但IN(x+1)的域限为x0。

Example 3
::例3

Evaluate: $\begin{aligned} \int \frac{4 x + 1}{4 x^{2} + 2 x + 1} d x \end{aligned}$
::评价:4x+14x2+2x+1dx

For $\begin{aligned} \int \frac{4 x + 1}{4 x^{2} + 2 x + 1} d x \end{aligned}$ , again try an see if $\begin{aligned} u \end{aligned}$ -substitution can work.
::对于 {% 4x+14x2+2x+1dx, 请再次尝试查看 u- 替代是否有效。

Let $\begin{aligned} u = 4 x^{2} + 2 x + 1 \end{aligned}$ ; then $\begin{aligned} d u = 8 x + 2 = 2 (4 x + 1) \end{aligned}$ .
::Letu=4x2+2x+1; 然后du=8x+2=2(4x+1)。

As you can see this substitution looks good because $\begin{aligned} d u = 2 (4 x + 1) \end{aligned}$ is a scaled version of the numerator. Then
::您可以看到这个替换看起来不错, 因为 du=2( 4x+1) 是分子的缩放版本。然后是 du= 2, 4x+1 。

$\begin{aligned} \int \frac{4 x + 1}{4 x^{2} + 2 x + 1} d x & = \int \frac{d u}{2 u} \\ = \frac{1}{2} \int \frac{d u}{u} \\ = \frac{1}{2} \ln | u | + C \\ = \frac{1}{2} \ln | 4 x^{2} + 2 x + 1 | + C \end{aligned}$

::@ 4x+14x2+2x+1dx#du2u=12 @duu=12lnu @C=12ln4x2+2x+1{C

Example 4
::例4

The metal tip of a pointer can be described as the rotation of the function $\begin{aligned} y = 0.25 \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} \end{aligned}$ inches from $\begin{aligned} 0 \leq x \leq 2 \end{aligned}$ inches about the $\begin{aligned} x \end{aligned}$ -axis. Find the surface area of the metal tip.
::指针的金属端可描述为函数 y=0.25 ex-e-xex+e-x英寸的旋转。从 X 轴的 0 x 2 英寸。查找金属端的表面区域。

Recall that the area, $\begin{aligned} S \end{aligned}$ , of the surface of revolution can be formulated as:
::回顾革命表面的S区可表述为:

$\begin{aligned} S = \int_{a}^{b} 2 π y d x = \int_{0}^{2} 2 π \cdot 0.25 \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} d x = \frac{π}{2} \int_{0}^{2} \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} d x . \end{aligned}$

::Sab2ydx0220.25ex-e-xex+e-xdx202ex-ex-xex+e-e-xdxxx。

If you stare at the integrand long enough, you eventually see that the numerator in the integrand is the derivative of the denominator, so that $\begin{aligned} u \end{aligned}$ -substitution will be helpful.
::如果你盯着一兆字节看足够长的时间, 你最终会看到一兆字节中的分子是分母的衍生物, 这样u替代会很有帮助。

Let $\begin{aligned} u = e^{x} + e^{- x} \end{aligned}$ ; then $\begin{aligned} d u = e^{x} - e^{- x} \end{aligned}$ , and the integral becomes
::Let u=ex+e-x; 然后du=ex-e-x, 整体成为

$\begin{aligned} S & = \frac{π}{2} \int_{0}^{2} \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} d x \\ = \frac{π}{2} \int_{2}^{(e^{2} + e^{- 2})} \frac{d u}{u} \\ = \frac{π}{2} \ln | u |_{2}^{(e^{2} + e^{- 2})} \\ = \frac{π}{2} (\ln (e^{2} + e^{- 2}) - \ln 2) \\ = \frac{π}{2} (\ln (\frac{e^{2} + e^{- 2}}{2})) \\ \approx 2.1 \end{aligned}$

::S202ex-e-xex+e-xdx%22(e2+e-2)、duu2ln22(e2+e-2)、2(e2+e-2)、2(e2+e-2-)-ln2)、2(l_(e2+e-22))2.1

The surface area of the metal tip is approximately $\begin{aligned} 2 {in}^{2} \end{aligned}$ .
::金属尖的表面面积约为2英寸2。

Example 5
::例5

Evaluate $\begin{aligned} \int_{0}^{1} \frac{5^{x} + 6^{x}}{7^{x}} d x \end{aligned}$ .
::=============================================================================================================================================== ===============================================================================================================================================================================================

To solve, rewrite each exponential in the integrand in terms of the common base $\begin{aligned} e \end{aligned}$ , before evaluating:
::要解答,在评估之前,以共同基数e 重写原数中的每个指数 :

$\begin{aligned} \int_{0}^{1} \frac{5^{x} + 6^{x}}{7^{x}} d x & = \int_{0}^{1} \frac{(e^{\ln 5})^{x} + (e^{\ln 6})^{x}}{(e^{\ln 7})^{x}} d x \\ = \int_{0}^{1} \frac{e^{x \ln 5} + e^{x \ln 6}}{e^{x \ln 7}} d x \\ = \int_{0}^{1} [e^{x \ln (\frac{5}{7})} + e^{x \ln (\frac{6}{7})}] d x \\ = {[\frac{e^{x \ln (\frac{5}{7})}}{\ln (\frac{5}{7})} + \frac{e^{x \ln (\frac{6}{7})}}{\ln (\frac{6}{7})}]}_{0}^{1} \\ = \frac{\frac{5}{7} - 1}{\ln (\frac{5}{7})} + \frac{\frac{6}{7} - 1}{\ln (\frac{6}{7})} \\ = \frac{- 2}{7 \ln (\frac{5}{7})} + \frac{- 1}{7 \ln (\frac{6}{7})} \end{aligned}$

::015x+6x7xxx01(eln5)x+(eln6)xx(eln7)xxxxx1Exln=5+exln7dx01[exln(57)+exln(67)dx=[exln(57)_(57)_(57)+exln(67)_(67)_(67)_(57)_(67)_(67)_(67)_(67)_(67)_(7)xxxxx+6x7x7xxxxx1xxxxxxxx_01#01[57(57)+exln}(67)]dx=[xln_(57)_(57)_(57)+exln_(57)+exln_(67)}01=571-1__00_(57_(57)_(67)_1}(67)

Review
::回顾

Evaluate the following integrals.
::评估以下综合体。

$\begin{aligned} \int 4 e^{x} d x \end{aligned}$
::4exdx

$\begin{aligned} \int 2 e^{2 x + 3} d x \end{aligned}$
::2e2x+3dx

$\begin{aligned} \int 5^{3 x + 2} d x \end{aligned}$
::53x+2dx

$\begin{aligned} \int (7 e^{- x} + 2) d x \end{aligned}$
:7e-x+2)dx

$\begin{aligned} \int e^{x} (1 + 3 e^{x})^{5} d x \end{aligned}$
::ex( 1+3ex) 5dx

$\begin{aligned} \int 5^{x} (1 - 5^{x}) (1 + 5^{x})^{6} d x \end{aligned}$
::5x(1-5x)(1+5x)6dx

$\begin{aligned} \int \frac{1}{e^{x}} d x \end{aligned}$
::1exdx

$\begin{aligned} \int \sqrt{e^{x}} d x \end{aligned}$
::exdx

$\begin{aligned} \int \frac{4 x - 3}{4 x^{2} - 6 x + 7} d x \end{aligned}$
::4x-34x2 -6x+7dx

$\begin{aligned} \int \frac{e^{x} + e^{- x}}{e^{x} - e^{- x}} d x \end{aligned}$
::* ex- e- xex- e- xdx

$\begin{aligned} \int_{0}^{e} \frac{d x}{x + e} \end{aligned}$
::0edxx+e

$\begin{aligned} \int_{- \ln 3}^{\ln 3} \frac{e^{x}}{e^{x} + 4} d x \end{aligned}$
::3ln3exex+4dx

$\begin{aligned} \int \frac{d x}{x \ln x} \end{aligned}$
::* dxxln *xxx * *xxxx *xxx *xxx *xxxx *xxxx *xxx *xxxx *xxxxx

$\begin{aligned} \int_{1}^{e^{\frac{π}{2}}} \frac{4 \cos (\ln x)}{x} d x \end{aligned}$
::===================================================================================================================================

$\begin{aligned} \int \frac{\ln x + 5}{x (\ln^{2} x + 4 \ln x + 4)} d x \end{aligned}$
::=================================================================================================================================================================================================

Hint: Break up the integrand into two parts and use $\begin{aligned} u \end{aligned}$ -substitution.
::提示:将原群分成两部分,使用替代手段。

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

Integrating Exponential and Logarithmic Functions ::集成指数函数和对数函数

Integrals Involving Exponential Functions ::涉及生态功能的综合系统

Integrals Involving Logarithmic Functions ::包含对对数函数的元件

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Review ::回顾

Review (Answers) ::回顾(答复)

Integrating Exponential and Logarithmic Functions
::集成指数函数和对数函数

Integrals Involving Exponential Functions
::涉及生态功能的综合系统

Integrals Involving Logarithmic Functions
::包含对对数函数的元件

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Example 5
::例5

Review
::回顾

Review (Answers)
::回顾(答复)