8.10 不当综合物:有中断的几代人

Section outline

• Select section General

  Collapse Expand

  General

  Collapse all Expand all

  • Select activity 新闻通告

    

    新闻通告 Forum
• Select section 不当综合器件: 与断裂相隔的元件

  Collapse Expand

  不当综合器件: 与断裂相隔的元件

  播放段落

  Evaluating some definite integrals without looking to determine whether the integral is an improper integral , can sometimes lead to an erroneous solution. As an example, try evaluating the integral $\begin{array}{r}\underset{0}{\overset{\pi }{\int }}{\sec }^{2}xdx\end{array}$  before proceeding further in this concept. Do you know why the integral is improper?
  ::评估某些明确的整体体而不考虑确定整体体是否不适当的整体体,有时可能导致错误的解决方案。 例如,在进一步推进这个概念之前,先尝试评估整体体 0sec2 xdx。 你知道整体体为何不合适吗?

  播放段落

  Integrands with Discontinuities
  ::与断裂状态相隔的曾几代人

  播放段落

  In this concept we continue the discussion of  improper integrals. Recall that the reason for the term improper is because these integrals either:
  ::在这一概念中,我们继续讨论不适当的整体体。

  播放段落
  1. include integration over or
     ::包括合并超过或
  播放段落
  2. the integrand may become infinite within the limits of integration.
     ::在融合的限度内,整数可能变得无限。

  播放段落

  In this concept we will look at case 2, where the integrand may be discontinuous within the limits of integration. This means that the integrand has a vertical asymptote (an infinite discontinuity) at the limit of integration, or at some point in the interval of integration.
  ::在这一概念中,我们将研究情况2, 原数在一体化的限度内可能是不连续的, 也就是说,原数在一体化的限度或一体化的间隔的某一点上有一个垂直的无连续性(无限的不连续性) 。

  播放段落

  For integrals with discontinuities, the following properties hold:
  ::对于有不连续性的组成部分,下列属性持有:

  播放段落
  • If $\begin{array}{r}f\end{array}$  is discontinuous at $\begin{array}{r}x=a\end{array}$  in the interval  $\begin{array}{r}[a,b]\end{array}$ , then the improper integral  $\begin{array}{r}\underset{a}{\overset{b}{\int }}f(x)dx\end{array}$ is defined as:
    $$\begin{array}{r}\underset{a}{\overset{b}{\int }}f(x)dx=\underset{l\to a^{+}}{lim}\underset{l}{\overset{b}{\int }}f(x)dx\end{array}$$
    
    ::如果 f 在 [a, b] 间距 x=a 时不连续, 则不适当的 {abf( x) dx 定义为 {abf( x) dx=limla* lbf( x) dx
  播放段落
  • If $\begin{array}{r}f\end{array}$  is discontinuous at  $\begin{array}{r}x=b\end{array}$ over the interval  $\begin{array}{r}[a,b]\end{array}$ , then the improper integral $\begin{array}{r}\underset{a}{\overset{b}{\int }}f(x)dx\end{array}$ is defined as:
    $$\begin{array}{r}\underset{a}{\overset{b}{\int }}f(x)dx=\underset{l\to b^{-}}{lim}\underset{a}{\overset{l}{\int }}f(x)dx.\end{array}$$
    
    ::如果 f 在 x=b 时间间隔[ a, b] 时不连续, 则不适当的 {abf( x) dx 被定义为 {abf( x) dx=limlbalf( x) dx 。
  播放段落
  • If the integrand  $\begin{array}{r}f\end{array}$  is discontinuous at $\begin{array}{r}x=c\end{array}$  over the interval  $\begin{array}{r}[a,b]\end{array}$  then the improper integral  $\begin{array}{r}\underset{a}{\overset{b}{\int }}f(x)dx\end{array}$ is defined as:
    ::如果在 x=c 时,正方位 f 在间隔[a,b] 上不连续,则不适当的构件 {abf(x)dx 被定义为:

  播放段落

  $$\begin{array}{r}\underset{a}{\overset{b}{\int }}f(x)dx=\underset{l\to c^{-}}{lim}\underset{a}{\overset{l}{\int }}f(x)dx+\underset{l\to c^{+}}{lim}\underset{l}{\overset{b}{\int }}f(x)dx.\end{array}$$

  
  ::{abf(x)dx=limlcalf(x)dx+limlc}lbf(x)dx。

  播放段落

  Recall from Chapter 5 in the Lesson on Definite Integrals that in order for the function  $\begin{array}{r}f\end{array}$  to be integrable, it must be bounded on the interval  $\begin{array}{r}[a,b]\end{array}$ . Otherwise, the function is not integrable and thus does not exist . For example, the integral
  ::回顾《关于缺陷综合体的经验教训》第5章第5节,为了使函数f不可识别,它必须受[a,b]间隔的约束,否则,函数不易识别,因此不存在。

  播放段落

  $$\begin{array}{r}\underset{0}{\overset{4}{\int }}\frac{dx}{x-1}\end{array}$$

  
  ::04dxx-1号

  播放段落

  develops an infinite discontinuity at  $\begin{array}{r}x=1\end{array}$  because the integrand approaches infinity at this point. However, it is continuous on the two intervals  $\begin{array}{r}[0,1)\end{array}$  and  $\begin{array}{r}(1,4]\end{array}$ . Looking at the integral more carefully, we may split the interval $\begin{array}{r}[0,4]\to [0,1)\cup (1,4]\end{array}$  and integrate between those two intervals to see if the integral converges .
  ::在 x=1 上形成无限不连续性, 原因是当时的整数法无限。 但是, 它在两个间隔[ 1 和(1 4) 上是连续的。 仔细看整数, 我们可能会将间隔[ 0 4] {[ 0 , 1 }\ {( 1 , 4 ) 和这两个间隔间进行整合, 以查看整数是否趋同 。

  播放段落

  $$\begin{array}{r}\underset{0}{\overset{4}{\int }}\frac{dx}{x-1}=\underset{0}{\overset{1}{\int }}\frac{dx}{x-1}+\underset{1}{\overset{4}{\int }}\frac{dx}{x-1}.\end{array}$$

  
  ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}

  播放段落

  We next evaluate each improper integral. Integrating the first integral on the right hand side,
  ::接下来我们评估每个不适当的整体体, 将第一个整体体结合到右手边,

  播放段落

  $$\begin{array}{rl}\underset{0}{\overset{1}{\int }}\frac{dx}{x-1}& =\underset{l\to 1^{-}}{lim}\underset{0}{\overset{1}{\int }}\frac{dx}{x-1}\\ & =\underset{l\to 1^{-}}{lim}[\ln |x-1|{]}_0^l\\ & =\underset{l\to 1^{-}}{lim}[\ln |l-1|-\ln |-1|]\\ & =-\mathrm{\infty }.\end{array}$$

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

  播放段落

  The integral diverges because  $\begin{array}{r}\ln (0)\end{array}$  is undefined, and thus there is no reason to evaluate the second integral. We conclude that the original integral diverges and has no finite value.
  ::整体差异是因为In(0)没有定义,因此没有理由评估第二个整体。 我们的结论是,最初的整体差异没有限制价值。

  播放段落

  Examples
  ::实例

  播放段落

  Example 1
  ::例1

  播放段落

  Earlier, you were asked to evaluate the improper integral  $\begin{array}{r}\underset{0}{\overset{\pi }{\int }}{\sec }^{2}xdx\end{array}$ . 
  ::早些时候,你被要求评估 不当的内装 0sec2xdx。

  播放段落

  Did you get that $\begin{array}{r}\underset{0}{\overset{\pi }{\int }}{\sec }^{2}xdx=\tan \pi -\tan 0=0\end{array}$ ? This is incorrect!
  ::你拿到那张#0sec2\\\ xdx=tan\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

  播放段落

  The integral $\begin{array}{r}\underset{0}{\overset{\pi }{\int }}{\sec }^{2}xdx\end{array}$  is improper because $\begin{array}{r}\sec x\end{array}$  is not defined at $\begin{array}{r}x=\frac{\pi }{2}\end{array}$ , and must be evaluated using the limit approach described in this concept. Doing so results in the following:
  ::构件 0sec2 xdx 是不当的, 因为 秒x 在 x2 上没有定义, 必须使用这个概念中描述的限值方法进行评估 。 这样做的结果如下:

  播放段落

  $$\begin{array}{r}\underset{0}{\overset{\pi }{\int }}{\sec }^{2}xdx=\underset{p\to -\frac{\pi }{2}}{lim}\underset{0}{\overset{p}{\int }}{\sec }^{2}xdx+\underset{p\to +\frac{\pi }{2}}{lim}\underset{p}{\overset{\pi }{\int }}{\sec }^{2}xdx=\underset{p\to -\frac{\pi }{2}}{lim}[\tan p]+\underset{p\to +\frac{\pi }{2}}{lim}[-\tan p].\end{array}$$

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

  播放段落

  But $\begin{array}{r}\underset{p\to -\frac{\pi }{2}}{lim}[\tan p]=\mathrm{\infty }\end{array}$  and diverges, and so does $\begin{array}{r}\underset{p\to +\frac{\pi }{2}}{lim}[-\tan p]\end{array}$ . So the integral diverges.
  ::但是瘸了,又裂了,又裂了, 也裂了。所以整体的裂了。

  播放段落

  Example 2
  ::例2

  播放段落

  Evaluate  $\begin{array}{r}\underset{1}{\overset{3}{\int }}\frac{dx}{\sqrt{x-1}}\end{array}$ .
  ::评估13dxx-1。

  播放段落

  $$\begin{array}{rl}\underset{1}{\overset{3}{\int }}\frac{dx}{\sqrt{x-1}}& =\underset{l\to 1^{+}}{lim}\underset{1}{\overset{3}{\int }}\frac{dx}{\sqrt{x-1}}\\ & =\underset{l\to 1^{+}}{lim}{\left[2\sqrt{x-1}\right]}_l^3\\ & =\underset{l\to 1^{+}}{lim}\left[2\sqrt{2}-2\sqrt{l-1}\right]\\ & =2\sqrt{2}.\end{array}$$

  
  ::13dxx-1=liml113dx-1=liml1+[2x-1-1]l3=liml1+[22-2l-1]=22。

  播放段落

  So the integral converges to  $\begin{array}{r}2\sqrt{2}\end{array}$ .
  ::因此,整体体会合到22。

  播放段落

  Example 3
  ::例3

  播放段落

  In Chapter 6 you learned to find the volume of a solid by revolving a curve. Let the curve be  $\begin{array}{r}y=x{e}^{-x},0\le x\le \mathrm{\infty }\end{array}$ , and revolve it about the  $\begin{array}{r}x\end{array}$ -axis. What is the volume of revolution?
  ::在第6章中,您学会通过旋转曲线找到固体的体积。让曲线为 y=xe-x,0x,然后围绕 X 轴旋转。革命的体积是什么?

  

  播放段落

  From the figure above, the area of the region to be revolved is given by  $\begin{array}{r}A=\pi y^2=\pi x^2{e}^{-2x}\end{array}$ . Thus the volume of the solid is
  ::从上图中,Ay2x2e-2xxxx给出了该区域转折地区。

  播放段落

  $$\begin{array}{r}V=\pi \underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^2{e}^{-2x}dx=\pi \underset{l\to \mathrm{\infty }}{lim}\underset{0}{\overset{l}{\int }}{x}^2{e}^{-2x}dx.\end{array}$$

  
  ::V0x2e -2xxxxxl=0lx2e -2xxxxx

  播放段落

  As you can see, we need to integrate by parts twice:
  ::如你所见 我们需要分两部分整合

  播放段落

  $$\begin{array}{rl}\int x^2{e}^{-2x}dx& =-\frac{x^2}{2}{e}^{-2x}+\int x{e}^{-2x}dx\\ & =-\frac{x^2}{2}{e}^{-2x}-\frac{x}{2}{e}^{-2x}-\frac{1}{4}{e}^{-2x}+C.\end{array}$$

  
  ::x2e-2xdxx22e-2xxxxxxxxxxxxxxxxxx22e-2x-x2e-2x-14e-2x+C。

  播放段落

  Thus
  ::因此,

  播放段落

  $$\begin{array}{rl}V& =\pi \underset{l\to \mathrm{\infty }}{lim}{\left[-\frac{x^2}{2}{e}^{-2x}-\frac{x}{2}{e}^{-2x}-\frac{1}{4}{e}^{-2x}\right]}_0^l\\ & =\pi \underset{l\to \mathrm{\infty }}{lim}{\left[\frac{2x^2+2x+1}{-4e^{2x}}\right]}_0^l\\ & =\pi \underset{l\to \mathrm{\infty }}{lim}\left[\frac{2l^2+2l+1}{-4e^{2l}}-\frac{1}{-4e^0}\right]\\ & =\pi \underset{l\to \mathrm{\infty }}{lim}\left[\frac{2l^2+2l+1}{4e^{2l}}+\frac{1}{4}\right].\end{array}$$

  
  ::Vliml[-x22e-2x-x2x-x2e-2x-14e-2x]0ll[2x2+2x+1+1-4e2x]0ll[2l2+2l+2l+1-4e2l-1-4e2l-4e0]l[2l2+2l+14e2l+14]。

  播放段落

  At this stage, we take the limit as  $\begin{array}{r}l\end{array}$  approaches infinity. Notice that the when you substitute infinity into the function, the denominator of the expression  $\begin{array}{r}\frac{2l^2+2l+1}{-4e^{2l}},\end{array}$  being an exponential function, will approach infinity at a much faster rate than will the numerator. Thus this expression will approach zero at infinity. Hence

  $$\begin{array}{r}V=\pi \left[0+\frac{1}{4}\right]=\frac{\pi }{4},\end{array}$$

  So the volume of the solid is  $\begin{array}{r}\frac{\pi }{4}\end{array}$ .
  ::在目前阶段, 我们以 I 接近无限的方式使用限制值。 请注意, 当您在函数中替换无限值时, 表达式 2l2+2l+1-4e2l 的分母, 是一个指数函数, 其无限度的接近速度将远快于该分子。 因此, 此表达式在无限度时将接近零 。 因此 V {[ 0+14] _4] 4, 所以固态的体积是 +4 。

  播放段落

  Example 4
  ::例4

  播放段落

  Evaluate  $\begin{array}{r}\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}\frac{dx}{e^x+e^{-x}}\end{array}$ .
  ::评估 dxex+e-x。

  播放段落

  This can be a tough integral! To simplify, rewrite the integrand as
  ::要简化, 将整数重写为

  播放段落

  $$\begin{array}{r}\frac{1}{e^x+e^{-x}}=\frac{1}{e^{-x}(e^{2x}+1)}=\frac{e^x}{e^{2x}+1}=\frac{e^x}{1+(e^x{)}^2}.\end{array}$$

  
  ::1ex+e-x=1e-x(e2x+1)=exe2x+1=ex1+(ex)2。

  播放段落

  Substitute into the integral:
  ::替换为集成件:

  播放段落

  $$\begin{array}{r}\int \frac{dx}{e^x+e^{-x}}=\int \frac{e^x}{1+(e^x{)}^2}dx.\end{array}$$

  
  ::dxex+e-xx1+(ex)2dx。

  播放段落

  Using  $\begin{array}{r}u\end{array}$ -substitution, let $\begin{array}{r}u=e^x,du=e^{x}dx.\end{array}$
  ::使用 u 替代, let u =ex, du =exdx 。

  播放段落

  $$\begin{array}{rl}\int \frac{dx}{e^x+e^{-x}}& =\int \frac{du}{1+u^2}\\ & ={\tan }^{-1}u+C\\ & ={\tan }^{-1}{e}^x+C.\end{array}$$

  
  ::dxex+e-xdu1+u2=tan-1u+C=tan-1ax+C。

  播放段落

  Returning to our integral with infinite limits, we split it into two regions. Choose as the split point the convenient  $\begin{array}{r}x=0.\end{array}$
  ::回到我们的有机体 无限的极限, 我们把它分成两个区域。 选择方便的 x=0 作为分割点 。

  播放段落

  $$\begin{array}{r}\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}\frac{dx}{e^x+e^{-x}}=\underset{-\mathrm{\infty }}{\overset{0}{\int }}\frac{dx}{e^x+e^{-x}}+\underset{0}{\overset{+\mathrm{\infty }}{\int }}\frac{dx}{e^x+e^{-x}}.\end{array}$$

  
  ::dxex+e -x0dxex+e -x0}dxex+e -x

  播放段落

  Taking each integral separately,
  ::将每个组成部分分别拆分,

  播放段落

  $$\begin{array}{rl}\underset{-\mathrm{\infty }}{\overset{0}{\int }}\frac{dx}{e^x+e^{-x}}& =\underset{l\to -\mathrm{\infty }}{lim}{\int }_l^0\frac{dx}{e^x+e^{-x}}\\ & =\underset{l\to -\mathrm{\infty }}{lim}{\left[{\tan }^{-1}{e}^x\right]}_l^0\\ & =\underset{l\to -\mathrm{\infty }}{lim}\left[{\tan }^{-1}{e}^0-{\tan }^{-1}{e}^l\right]\\ & =\frac{\pi }{4}-0\\ & =\frac{\pi }{4}.\end{array}$$

  
  ::0dxex+e-x=limll0dxex+e-x=liml[tan-1ex]l0=liml[tan-1e0-tan-1_1el]4-04。

  播放段落

  Similarly,
  ::同样,

  播放段落

  $$\begin{array}{rl}\underset{0}{\overset{+\mathrm{\infty }}{\int }}\frac{dx}{e^x+e^{-x}}& =\underset{l\to \mathrm{\infty }}{lim}\underset{0}{\overset{1}{\int }}\frac{dx}{e^x+e^{-x}}\\ & =\underset{l\to \mathrm{\infty }}{lim}{\left[{\tan }^{-1}{e}^x\right]}_0^l\\ & =\underset{l\to \mathrm{\infty }}{lim}\left[{\tan }^{-1}{e}^l-{\tan }^{-1}1\right]\\ & =\frac{\pi }{2}-\frac{\pi }{4}\\ & =\frac{\pi }{4}.\end{array}$$

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

  播放段落

  Thus the integral converges to
  ::因此,整体体汇合到

  播放段落

  $$\begin{array}{r}\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}\frac{dx}{e^x+e^{-x}}=\frac{\pi }{4}+\frac{\pi }{4}=\frac{\pi }{2}.\end{array}$$

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

  播放段落

  Example 5
  ::例5

  播放段落

  Evaluate the integral $\begin{array}{r}\underset{1}{\overset{2}{\int }}\frac{1}{x\sqrt{x^2-1}}dx\end{array}$ .

  $$\begin{array}{rl}\underset{1}{\overset{2}{\int }}\frac{1}{x\sqrt{x^2-1}}dx& =\underset{t\to 1^{+}}{lim}\underset{t}{\overset{2}{\int }}\frac{1}{x\sqrt{x^2-1}}dx\\ & =\underset{t\to 1^{+}}{lim}[arc\sec x{]}_t^2\\ & =arc\sec 2-\underset{t\to 1^{+}}{lim}[arc\sec t]\\ & =arc\cos \left(\frac{1}{2}\right)-\underset{t\to 1^{+}}{lim}\left[arc\cos \frac{1}{t}\right]\\ & =\frac{\pi }{3}-0\end{array}$$

  
  ::-=============================================================================================================================================================================================================================================================================================================================================================================================================================================================================================================================

  播放段落

  Therefore, $\begin{array}{r}\underset{1}{\overset{2}{\int }}\frac{1}{x\sqrt{x^2-1}}dx=\frac{\pi }{3}\end{array}$ .
  ::因此,112x22-1dx_3。

  播放段落

  Review
  ::回顾

  播放段落
  1. Determine whether the following integrals are improper. If so, explain why.
     播放段落
     1. $\begin{array}{r}\underset{5}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{(x-2{)}^2}\end{array}$
        ::5dx(x-2)2
     播放段落
     2. $\begin{array}{r}\underset{2}{\overset{5}{\int }}\frac{dx}{(x-2{)}^2}\end{array}$
        ::25dx(x-2)2
     播放段落
     3. $\begin{array}{r}\underset{3}{\overset{5}{\int }}\frac{dx}{(x-2{)}^2}\end{array}$
        ::35dx(x-2)2
     播放段落
     4. $\begin{array}{r}\underset{4}{\overset{10}{\int }}\frac{1}{\sqrt{x-2}}dx\end{array}$
        ::4101x-2dx
     播放段落
     5. $\begin{array}{r}\underset{0}{\overset{\frac{\pi }{4}}{\int }}\tan 4xdx\end{array}$
        ::04tan4xxx
     播放段落
     6. $\begin{array}{r}\underset{0}{\overset{10}{\int }}\frac{x+4}{x^2-2x-15}dx\end{array}$
        ::010x+4x2-2x-15dx
     
     ::确定以下的内装件是否不合适 。 如果是的话, 请解释原因 。 @ @ 5dx( x-2) 2 = 25dx( x-2) 2 = 35dx( x-2) 2 = 35dx( x-2) 2 = 4101x-2dx = 04tan* 4xdx = 010x+4x2- 2x- 15dx

  播放段落

   For #2-15, evaluate the integral or state that it diverges.
  ::对于#2-15, 评估它所不同的整体或状态。

  播放段落
  2. $\begin{array}{r}\underset{0}{\overset{4}{\int }}\frac{1}{\sqrt{x}}dx\end{array}$
     ::041xxx
  播放段落
  3. $\begin{array}{r}\underset{1}{\overset{8}{\int }}\frac{1}{\sqrt{8-x}}dx\end{array}$
     ::1818-xdx
  播放段落
  4. $\begin{array}{r}\underset{0}{\overset{4}{\int }}\frac{e^{\sqrt{x}}}{\sqrt{x}}dx\end{array}$
     ::04xxxxxxxx
  播放段落
  5. $\begin{array}{r}\underset{3}{\overset{5}{\int }}\frac{1}{(x-3{)}^4}dx\end{array}$
     ::351(x-3-3)4dx
  播放段落
  6. $\begin{array}{r}\underset{0}{\overset{\frac{\pi }{2}}{\int }}\sec xdx\end{array}$
     ::02secxdx
  播放段落
  7. $\begin{array}{r}\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{1-x^2}}dx\end{array}$
     ::011-x2dx
  播放段落
  8. $\begin{array}{r}\underset{0}{\overset{3}{\int }}\frac{x}{\sqrt{9-x^2}}dx\end{array}$
     ::03x9 - x2dx
  播放段落
  9. $\begin{array}{r}\underset{0}{\overset{5}{\int }}\frac{5x+1}{x^2-2x-3}dx\end{array}$
     ::055x+1x2-2x-3dx
  播放段落
  10. $\begin{array}{r}\underset{0}{\overset{\sqrt{\pi }}{\int }}x{\tan }^2(x^2){\sec }^2(x^2)dx\end{array}$
      ::0xtan2(x2)seec2(x2)dx
  播放段落
  11. $\begin{array}{r}\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{1}{x^2}dx\end{array}$
      ::01x2dx
  播放段落
  12. $\begin{array}{r}\underset{-1}{\overset{8}{\int }}\frac{1}{\sqrt[3]{x}}dx\end{array}$
      ::181x3dx
  播放段落
  13. $\begin{array}{r}\underset{0}{\overset{1}{\int }}\ln xdx\end{array}$
      ::01lnxdx
  播放段落
  14. $\begin{array}{r}\underset{1}{\overset{2}{\int }}\frac{1}{x\ln x}dx\end{array}$
      ::121xlnxdx
  播放段落
  15. $\begin{array}{r}\underset{1}{\overset{e}{\int }}\frac{1}{x(\ln x{)}^2}dx\end{array}$
      ::1e1x( lnx) 2dx

  播放段落

  Review (Answers)
  ::回顾(答复)

  播放段落

  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 。