5.3 无限期融合:变数变化

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• 选择节无限整合:变量变化

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  无限整合:变量变化

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  In a previous concept you learned how to find the derivative of a composite function using the . An obvious question in discussing antiderivatives: Is there some similar rule or method that can help in finding the of a composite function? The answer is yes! By performing a change of variable, or variable substitution, the integrand can sometimes be made so that it is easier to evaluate, or find an antiderivative.
  ::在先前的概念中,您学会了如何使用 。在讨论抗降解剂时,一个显而易见的问题:是否有类似的规则或方法可以帮助找到复合函数?答案是肯定的!通过改变变数或变数替代,有时可以使整数变得更容易评估或找到抗降解剂。

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

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  Let’s illustrate the problem and solution.
  ::让我们来说明问题和解决办法。

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  Suppose we needed to compute the following integral:
  ::假设我们需要计算以下组成部分:

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  $\begin{align*}\int[2x + 1]^2 dx\end{align*}$
  ::[2x+1] 2dx

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  We are looking for an antiderivative $\begin{align*}F\end{align*}$ of $\begin{align*}f\end{align*}$ , and can put together a solution by either of the following ways:
  ::我们正在寻找F的抗降解F,我们可以通过下列两种方式提出解决办法:

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  • Expand the integrand into the polynomial $\begin{align*}f(x) = 4x^2 + 4x + 1\end{align*}$ and use the Sum Rule for evaluating an integrand to obtain $\begin{align*}F(x) = \frac{4}{3}x^3 + \frac{4}{2}x^2 + x + C\end{align*}$ ; or
    ::将整数扩大为多位 f( x) = 4x2+4x+1, 并使用 Sum 规则评估整数以获取 F( x) = 43x3+42x2+xC;或
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  • Use our knowledge of differentiation to put together the function  $\begin{align*}F\end{align*}$ such that $\begin{align*}F(x) = \frac{1}{3 \cdot 2}[2x + 1]^3 + C\end{align*}$ .
    ::利用我们的差异知识将功能F(x)=132[2x+1]3+C结合起来。

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  Let’s consider a different approach. We recognize the function $\begin{align*}f(x) = [2x + 1]^2\end{align*}$ as a composite function $\begin{align*}f \circ g = f(g(x))\end{align*}$ , with $\begin{align*}f(x) = x^2\end{align*}$ and $\begin{align*}g(x) = 2x + 1\end{align*}$ . We know how to find the antiderivative of $\begin{align*}f(x) = x^2\end{align*}$ . But how do we factor in $\begin{align*}g(x) = 2x + 1\end{align*}$ ?
  ::让我们考虑另一种方法。 我们确认函数 f( x) =[ 2x+1] 。 我们确认函数 f( x) =[ 2x+1] 是一个复合函数 fg= f( g( x) ) , 与 f( x) =x2 和 g( x) = 2x+1. 。 我们知道如何找到 f( x) =x2 的抗衍生物。 但我们如何以 g( x) = 2x+1 计数 ?

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  1. Make the substitution $\begin{align*}u = 2x + 1\end{align*}$
     ::替换 u=2x+1
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  2. Then $\begin{align*}\frac{du}{dx} = 2\end{align*}$ , or $\begin{align*}du = 2dx\end{align*}$ . This means $\begin{align*}dx = \frac{du}{2}\end{align*}$ .
     ::然后是 dudx=2, 或 du=2dx。 这意味着 dx=du2 。
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  3. Now the integral can be represented as $\begin{align*}\int [2x + 1]^2 dx = \int u^2 \frac{du}{2} = \frac{1}{2}\int{u^2}du\end{align*}$ , which we know how to integrate.
     ::整体体现在可以以 @[2x+1]2dx@u2du2=12\u2表示, 我们知道如何整合。
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  4. The new integral is evaluated to find the antiderivative $\begin{align*}F(u) = \frac{1}{2}\int u^2 du = \frac{1}{3 \cdot 2}u^3 + C\end{align*}$ .
     ::对新的整体体进行评估,以找到F(u)=12u2du=132u3+C的抗降解性F(u)=12u2du=132u3+C。
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  5. Finally, we can return to using the variable $\begin{align*}x\end{align*}$ to get $\begin{align*}F(x) = \frac{1}{3 \cdot 2}[2x + 1]^3 + C\end{align*}$ .
     ::最后,我们可以回到使用变量 x 来获取 F(x) = 13.2[2x+1]3+C 。

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  Therefore,
  ::因此,

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  $\begin{align*}\int [2x + 1]^2 dx =\frac{1}{3 \cdot 2}[2x + 1]^3 + C.\end{align*}$
  ::[2x+1]2dx=13}2[2x+1]3+C。

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  The technique we have used is called Integration by Substitution , a technique that can often simplify the integrand considerably and allow us to recognize an antiderivative.
  ::我们使用的技术称为 " 替代融合 " ,这种技术往往可以大大简化原状,使我们认识到一种抗降解剂。

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  Sometimes the integrand looks more complex, and requires more effort to recognize how to simplify.
  ::有时,一万格朗看起来更复杂,需要更加努力地认识到如何简化。

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  Suppose we needed to compute the following integral:
  ::假设我们需要计算以下组成部分:

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  $\begin{align*}\int 3x^2\sqrt{1+x^3} dx\end{align*}$
  ::3x21+x3dx

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  The integrand is the product of $\begin{align*}3x^2\end{align*}$ and a composite function $\begin{align*}f \circ g = f(g(x))\end{align*}$ , with $\begin{align*}f(x) = \sqrt{x}\end{align*}$ and $\begin{align*}g(x) = 1+x^3\end{align*}$ .
  ::Integrand 是 3x2 和 f(x) 和 g(x) = 1+x3 的复合函数 fg=f(g(x)x) 和 g(x) = 1+x3 的产物。

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  Let’s try to use the substitution approach and see what happens.
  ::让我们尝试使用替代方法, 看看会发生什么。

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  1. Let $\begin{align*}u=1+x^3\end{align*}$ .
     ::Letu = 1+x3 。
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  2. Differentiate both sides so $\begin{align*}du = 3x^2 dx\end{align*}$ ; we recognize that  $\begin{align*}3x^2\end{align*}$ is $\begin{align*}g^{\prime}(x)\end{align*}$ , one of the factors in the integrand, and this fact is very helpful.
     ::将两边区分为 du= 3x2dx; 我们认识到, 3x2 是g*(x), 这是整数的一个因素, 这一事实非常有用。

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  Note that integral is of the form  $\begin{align*}\int f(g(x)) \cdot g^{\prime}(x)dx= \int F^{\prime}(g(x)) \cdot g^{\prime}(x)dx\end{align*}$ .
  ::注意,整件为表 {f(g(x))}}}{g}}{g}}}}(x)dx}}{F}}(g(x)}}}}}}(x)dxx。

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  3. Change the original integral in  $\begin{align*}x\end{align*}$ to an integral in $\begin{align*}u\end{align*}$ :
     ::将x的原积分修改为 u 的积分:

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  $\begin{align*}\int \sqrt{1+x^3} \cdot 3x^2 dx = \int \sqrt{u} du\end{align*}$ , where $\begin{align*}u = 1+x^3\end{align*}$ and $\begin{align*}du = 3x^2 dx\end{align*}$ .
  ::+1+x3=3x2dx=udu,其中u=1+x3和du=3x2dx。

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  4. Integrate with respect to $\begin{align*}u\end{align*}$ :
     ::结合u:

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  $\begin{align*}\int \sqrt{u} du= \int u^{\frac{1}{2}} du = \frac{2}{3}u^{\frac{3}{2}}+C\end{align*}$ .
  ::u12du=23u32+C。

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  5. Now change the answer back to $\begin{align*}x\end{align*}$ :
     ::现在将回答修改为 x:

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  $\begin{align*}\frac{2}{3}u^{\frac{3}{2}} + C = \frac{2}{3}(1 + x^3)^{\frac{3}{2}} + C\end{align*}$
  ::2332+C=23(1+x3)32+C

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  Substitution is a very powerful method for solving a variety of problems.
  ::替代是解决各种问题的非常有力的方法。

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  Compute the following indefinite integral:
  ::计算以下不定期整体部分:

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  $\begin{align*}\int x^2 e^{x^3} dx.\end{align*}$
  ::x2ex3dx.

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  We note that the current problem is not of the form $\begin{align*}\int F^{\prime}(g(x)) \cdot g^{\prime}(x)dx\end{align*}$ , like the last example, but that the derivative of  $\begin{align*}x^3\end{align*}$ is $\begin{align*}3x^2\end{align*}$ . This gives a clue to what the substitution should be.
  ::我们注意到,当前的问题不是像最后一个例子那样的表 {F}(g(x))}}}(x)}}(x)dx,而是 x3 的衍生物是 3x2 。 这为替换应该是什么提供了线索。

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  Let $\begin{align*}u=x^3\end{align*}$ .
  ::Letu =x3 。

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  Then $\begin{align*}du = 3{x^2}dx\end{align*}$ which means $\begin{align*}\frac{1}{3}du = {x^2}dx.\end{align*}$
  ::然后是DU=3x2dx,这意味着13du=x2dx。

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  Now we can change the original integral in  $\begin{align*}x\end{align*}$ to an integral in  $\begin{align*}u\end{align*}$ and integrate:
  ::现在,我们可以将x的原有机体改为u的有机体,并整合:

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  $\begin{align*}\int x^2 e^{x^3} dx= \int e^u \left(\frac{1}{3} du\right) =\frac{1}{3} \int e^u du= \frac{1}{3} e^u+C.\end{align*}$
  ::X2ex3dxeu(13du)=13eudu=13eu+C。

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  Changing back to $\begin{align*}x\end{align*}$ , we have
  ::回换到 x, 我们发现

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  $\begin{align*}\int x^2 e^{x^3} dx=\frac{1}{3}e^{x^3}+C.\end{align*}$
  ::X2ex3dx=13ex3+C。

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  Examples
  ::实例

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  Example 1
  ::例1

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  Suppose we needed to compute the following integral:
  ::假设我们需要计算以下组成部分:

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  $\begin{align*}\int \frac{x \sin \left(\sqrt{3x^2 + 1} \right)}{\sqrt{3x^2 +1}}dx\end{align*}$
  ::xsin( 3x2+1) {3x2+1} 3x2+1dx

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  We notice that the sine function in the integrand is a composite function $\begin{align*}f \circ g = f(g(x))\end{align*}$ , with $\begin{align*}f(x) = \sin x\end{align*}$ and $\begin{align*}g(x) = \sqrt{3x^2+ 1}\end{align*}$ .
  ::我们注意到,正弦函数在整数中是 fg=f(g(x)),与 f(x)=sinx 和 g(x)=3x2+1 的复合函数 fg=f(g(x)x)), 与 f(x)=sinx 和 g(x)=3x2+1 。

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  Let’s try to use the substitution approach and see what happens.
  ::让我们尝试使用替代方法, 看看会发生什么。

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  1. Let $\begin{align*}u = \sqrt{3x^2+ 1}\end{align*}$ .
     ::让我们% 3x2+1 。
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  2. Differentiate both sides so $\begin{align*}du = \frac{6x}{2 \sqrt{3x^2+1}}dx\end{align*}$ ; we recognize that $\begin{align*}\frac{du}{3} = \frac{x}{\sqrt{3x^2+1}}dx\end{align*}$ is $\begin{align*}\frac{1}{3}g^{\prime}(x)\end{align*}$ , one of the factors in the integrand, and this fact is very helpful.
     ::将两边区分为 du= 6x2}3x2+1dx; 我们认识到, du3=x}3x2+1dx是 13g_(x), 这是整数中的因素之一, 这一事实非常有用 。

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  Note that integral is of the form $\begin{align*}\int f(g(x)) \cdot g^{\prime}(x)dx=\int F^{\prime}(g(x)) \cdot g^{\prime}(x)dx\end{align*}$ .
  ::注意,整件为表 {f(g(x))}}}{g}}{g}}}}(x)dx}}{F}}(g(x)}}}}}}(x)dxx。

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  3. Change the original integral in $\begin{align*}x\end{align*}$ to an integral in $\begin{align*}u\end{align*}$ :
     ::将x的原积分修改为 u 的积分:

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  $\begin{align*}\int \frac{x \sin \sqrt {3x^2+1}}{\sqrt{3x^2+1}}dx = \frac{1}{3}\int \sin u \ du\end{align*}$
  ::3x2+1x3x2+1dx=13sinu

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  4. Integrate with respect to $\begin{align*}u\end{align*}$ :
     ::结合u:

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  $\begin{align*}\frac{1}{3}\int \sin u \ du = -\frac{1}{3}\cos u + C\end{align*}$
  ::1313cosu+C 13sinudu13cosu+C

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  5. Now change the answer back to  $\begin{align*}x\end{align*}$ :
     ::现在将回答修改为 x:

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  $\begin{align*}\int \frac{x \sin \sqrt{3x^2 +1}}{\sqrt{3x^2+1}}dx = -\frac{1}{3}\cos \sqrt {3{x^2} + 1} + C\end{align*}$
  ::Xsin3x2+1}3x2+1x2+1dx13cos3x2+1+C

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  Review
  ::回顾

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  Compute the integrals:
  ::计算积分 :

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  1. $\begin{align*}\int (3x + 8)^{11}dx\end{align*}$
     : 3x+8) 11dx
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  2. $\begin{align*}\int 5(5x - 3)^3 dx\end{align*}$
     ::5(5x-3)3dx
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  3. $\begin{align*}\int \sin 3x dx\end{align*}$
     ::辛3xxx
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  4. $\begin{align*}\int 5 \cos \frac{\pi }{2}xdx\end{align*}$
     ::5cos%2xxxx
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  5. $\begin{align*}\int x(2x^2 + 1)^2 dx\end{align*}$
     ::*x( 2x2+1) 2dx
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  6. $\begin{align*}\int (2x + 1)(x^2 + x - 1)^4 dx\end{align*}$
     : 2x+1)( x2+x- 1) 4dx
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  7. $\begin{align*}\int 4x^3 (7x^4 + 6)^5 dx\end{align*}$
     ::4x3( 7x4+6) 5dx
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  8. $\begin{align*}\int x^5 (2x^6 + 7)^3 dx\end{align*}$
     ::x5( 2x6+7) 3dx
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  9. $\begin{align*}\int \sec^2 x \tan xdx\end{align*}$
     ::
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  10. $\begin{align*}\int \frac{x}{\sqrt{2x + 1}} dx\end{align*}$
      ::x2x+1dx
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  11. $\begin{align*}\int \frac{x^2}{(5x^3+3)^{\frac{3}{2}}} dx\end{align*}$
      ::x2( 5x3+3) 32dx
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  12. $\begin{align*}\int x^3 \sqrt{1 - x^2} dx\end{align*}$
      ::x31 -x2dx
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  13. $\begin{align*}\int x^2 \sqrt{x^3+9} dx\end{align*}$
      ::x2x3+9dx
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  14. $\begin{align*}\int \left(x \cdot e^{x^2}\right) dx\end{align*}$
      :xex2)dx
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  15. $\begin{align*}\int \left(\frac{1}{x^2} \cdot e^{\frac{1}{x}}\right) dx\end{align*}$
      : 1x2e1x) dx

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  Review (Answers)
  ::回顾(答复)

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  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 。