2.2 Integration durch Substitution
Aus Online Mathematik Brückenkurs 2
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- | {{Vald flik|[[2.2 Variabelsubstitution| | + | {{Vald flik|[[2.2 Variabelsubstitution|Theory]]}} |
- | {{Ej vald flik|[[2.2 Övningar| | + | {{Ej vald flik|[[2.2 Övningar|Exercises]]}} |
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{{Info| | {{Info| | ||
- | ''' | + | '''Content:''' |
- | * | + | * Integration by substitution |
}} | }} | ||
{{Info| | {{Info| | ||
- | ''' | + | '''Learning outcomes:''' |
- | + | After this section, you will have learned to: | |
- | * | + | * Understand the derivation of the formula for variable substitution . |
- | * | + | * Solve easier integration problems that require rewriting and / or substitution in one of the steps. |
- | * | + | * Know how the limits of integration are to be changed after a variable substitution. |
- | * | + | * Know when variable substitution is allowed. |
}} | }} | ||
- | == | + | == Variable substitution == |
- | + | When you cannot directly determine a primitive function by using the usual rules of differentiation ”in the opposite direction”, method, other methods or techniques are needed . One such is ''variable substitution'', which can be said to be based on the rule for the differentiation of composite functions — the so-called ''chain rule''. | |
- | + | The chain rule <math>\ \frac{d}{dx}f(u(x)) = f^{\,\prime} (u(x)) \cdot u'(x)\ </math> can be written in integral form as | |
{{Fristående formel||<math>\int f^{\,\prime}(u(x)) \cdot u'(x) \, dx = f(u(x)) + C</math>}} | {{Fristående formel||<math>\int f^{\,\prime}(u(x)) \cdot u'(x) \, dx = f(u(x)) + C</math>}} | ||
- | + | or, | |
<div class="regel"> | <div class="regel"> | ||
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</div> | </div> | ||
- | + | where ''F'' is a primitive function of ''f''. We compare this with the formula | |
{{Fristående formel||<math>\int f(u) \, du = F(u) + C\,\mbox{,}</math>}} | {{Fristående formel||<math>\int f(u) \, du = F(u) + C\,\mbox{,}</math>}} | ||
- | + | We can see that we have replaced the term <math>u(x)</math> with variable <math>u</math> and the <math>u'(x)\, dx</math> with <math>du</math>. One thus can transform the more complicated integrand <math>f(u(x)) \cdot u'(x)</math> (with <math>x</math> as the variable) to the, let us hope, easier <math>f(u)</math> (with the <math>u</math> as the variable). The method is called variable substitution and can be used when the integrand can be written in the form <math>f(u(x)) \cdot u'(x)</math>. | |
- | '' | + | ''Note 1'' The method is based on the assumption that all the conditions for integration are satisfied; that is <math>u(x)</math> is differentiable in the interval in question, and that <math>f</math> is continuous for all values of <math>u</math> in the range, that is for all the values that <math>u</math> can take on in the interval. |
- | '' | + | '' Note 2'' Replacing <math>u'(x) \, dx</math> with <math>du</math> also may be justified by studying the transition from the increment ratio to the derivative: |
{{Fristående formel||<math>\lim_{\Delta x \to 0} \frac{\Delta u}{\Delta x} = \frac{du}{dx} = u'(x)\,\mbox{,}</math>}} | {{Fristående formel||<math>\lim_{\Delta x \to 0} \frac{\Delta u}{\Delta x} = \frac{du}{dx} = u'(x)\,\mbox{,}</math>}} | ||
- | + | which, as <math>\Delta x</math> goes towards zero can be considered as a formal transition between variables | |
{{Fristående formel||<math>\Delta u \approx u'(x) \Delta x \quad \to \quad du = u'(x) \, dx\,\mbox{,}</math>}} | {{Fristående formel||<math>\Delta u \approx u'(x) \Delta x \quad \to \quad du = u'(x) \, dx\,\mbox{,}</math>}} | ||
- | + | ie., a small change, <math>dx</math>, in the variable <math>x</math> gives rise to an approximate change <math>u'(x)\,dx</math> in the variable <math>u</math>. | |
<div class="exempel"> | <div class="exempel"> | ||
- | ''' | + | ''' Example 1''' |
- | + | Determine the integral<math>\ \int 2 x\, e^{x^2} \, dx</math>. | |
<br> | <br> | ||
<br> | <br> | ||
- | + | If one puts <math>u(x)= x^2</math>, one gets <math>u'(x)= 2x</math>. The variable substitution replaces <math>e^{x^2}</math> with <math>e^u</math> and <math>u'(x)\,dx</math>, i.e. <math>2x\,dx</math>, with <math>du</math> | |
{{Fristående formel||<math> \int 2 x\,e^{x^2} \, dx = \int e^{x^2} \cdot 2x \, dx = \int e^u \, du = e^u + C = e^{x^2} + C\,\mbox{.}</math>}} | {{Fristående formel||<math> \int 2 x\,e^{x^2} \, dx = \int e^{x^2} \cdot 2x \, dx = \int e^u \, du = e^u + C = e^{x^2} + C\,\mbox{.}</math>}} | ||
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<div class="exempel"> | <div class="exempel"> | ||
- | ''' | + | ''' Example 2''' |
- | + | Determine the integral <math>\ \int (x^3 + 1)^3 \cdot x^2 \, dx</math>. | |
<br> | <br> | ||
<br> | <br> | ||
- | + | Put <math>u=x^3 + 1</math>. This means <math>u'=3x^2</math>, or <math>du= 3x^2\, dx</math>, and | |
{{Fristående formel||<math>\begin{align*}\int (x^3 + 1)^3 x^2 \, dx &= \int \frac{ (x^3 + 1)^3}{3} \cdot 3x^2\, dx = \int \frac{u^3}{3}\, du\\[4pt] &= \frac{u^4}{12} + C = \frac{1}{12} (x^3 + 1)^4 + C\,\mbox{.}\end{align*}</math>}} | {{Fristående formel||<math>\begin{align*}\int (x^3 + 1)^3 x^2 \, dx &= \int \frac{ (x^3 + 1)^3}{3} \cdot 3x^2\, dx = \int \frac{u^3}{3}\, du\\[4pt] &= \frac{u^4}{12} + C = \frac{1}{12} (x^3 + 1)^4 + C\,\mbox{.}\end{align*}</math>}} | ||
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<div class="exempel"> | <div class="exempel"> | ||
- | ''' | + | ''' Example 3''' |
- | + | Determine the integral <math>\ \int \tan x \, dx\,\mbox{,}\ \ </math> where <math>-\pi/2 < x < \pi/2</math>. | |
<br> | <br> | ||
<br> | <br> | ||
- | + | After rewriting <math>\tan x</math> as <math>\sin x/\cos x</math> we substitute <math>u=\cos x</math>, | |
{{Fristående formel||<math>\begin{align*}\int \tan x \, dx &= \int \frac{\sin x}{\cos x} \, dx = \left[\,\begin{align*} u &= \cos x\\ u' &= - \sin x\\ du &= - \sin x \, dx \end{align*}\,\right]\\[4pt] &= \int -\frac{1}{u}\, du = - \ln |u| +C = -\ln |\cos x| + C\,\mbox{.}\end{align*}</math>}} | {{Fristående formel||<math>\begin{align*}\int \tan x \, dx &= \int \frac{\sin x}{\cos x} \, dx = \left[\,\begin{align*} u &= \cos x\\ u' &= - \sin x\\ du &= - \sin x \, dx \end{align*}\,\right]\\[4pt] &= \int -\frac{1}{u}\, du = - \ln |u| +C = -\ln |\cos x| + C\,\mbox{.}\end{align*}</math>}} | ||
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- | == | + | == The limits of integration during variable substitution. == |
- | + | ||
- | + | ||
+ | When calculating definite integrals, such as an area, one can go about using variable substitution in two ways. Either one can calculate the integral as usual and then switch back to the original variable and insert the original limits of integration. Alternatively one can change the limits of integration simultaneously with the variable substitution. The two methods are illustrated in the following example. | ||
<div class="exempel"> | <div class="exempel"> | ||
- | ''' | + | ''' Example 4''' |
- | + | Determine the integral <math>\ \int_{0}^{2} \frac{e^x}{1 + e^x} \, dx</math>. | |
- | '' | + | '' Method 1'' |
- | + | Put <math>u=e^x</math> which gives that <math>u'= e^x</math> and <math>du= e^x\,dx</math> | |
{{Fristående formel||<math>\begin{align*}\int_{0}^{2} \frac{e^x}{1 + e^x} \, dx &= \int_{x=0}^{\,x=2} \frac{1}{1 + u} \, du = \Bigl[\,\ln |1+ u |\,\Bigr]_{x=0}^{x=2} = \Bigl[\,\ln (1+ e^x)\,\Bigr]_{0}^{2}\\[4pt] &= \ln (1+ e^2) - \ln 2 = \ln \frac{1+ e^2}{2}\,\mbox {.}\end{align*}</math>}} | {{Fristående formel||<math>\begin{align*}\int_{0}^{2} \frac{e^x}{1 + e^x} \, dx &= \int_{x=0}^{\,x=2} \frac{1}{1 + u} \, du = \Bigl[\,\ln |1+ u |\,\Bigr]_{x=0}^{x=2} = \Bigl[\,\ln (1+ e^x)\,\Bigr]_{0}^{2}\\[4pt] &= \ln (1+ e^2) - \ln 2 = \ln \frac{1+ e^2}{2}\,\mbox {.}\end{align*}</math>}} | ||
- | + | Note that the limits of integration must be written in the form <math>x = 0</math> and <math>x = 2</math> when the variable of integration is not <math>x</math>. it is wrong to write | |
{{Fristående formel||<math>\int_{0}^{2} \frac{e^x}{1 + e^x} \, dx = \int_{0}^{2} \frac{1}{1 + u} \, du \quad \text{ osv.}</math>}} | {{Fristående formel||<math>\int_{0}^{2} \frac{e^x}{1 + e^x} \, dx = \int_{0}^{2} \frac{1}{1 + u} \, du \quad \text{ osv.}</math>}} | ||
- | '' | + | '' Method 2'' |
- | + | Put <math>u=e^x</math> which gives that <math>u'= e^x</math> and <math>du= e^x\, dx</math>. The limit of integration <math>x=0</math> is equivalent to <math>u=e^0 = 1</math> and <math>x=2</math> is equivalent to <math>u=e^2</math> | |
{{Fristående formel||<math>\int_{0}^{2} \frac{e^x}{1 + e^x} \, dx = \int_{1}^{\,e^2} \frac{1}{1 + u} \, du = \Bigl[\,\ln |1+ u |\,\Bigr]_{1}^{e^2} = \ln (1+ e^2) - \ln 2 = \ln\frac{1+ e^2}{2}\,\mbox{.}</math>}} | {{Fristående formel||<math>\int_{0}^{2} \frac{e^x}{1 + e^x} \, dx = \int_{1}^{\,e^2} \frac{1}{1 + u} \, du = \Bigl[\,\ln |1+ u |\,\Bigr]_{1}^{e^2} = \ln (1+ e^2) - \ln 2 = \ln\frac{1+ e^2}{2}\,\mbox{.}</math>}} | ||
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<div class="exempel"> | <div class="exempel"> | ||
- | ''' | + | ''' Example 5''' |
- | + | Determine the integral <math> \ \int_{0}^{\pi/2} \sin^3 x\,\cos x \, dx</math>. | |
<br> | <br> | ||
<br> | <br> | ||
- | + | The substitution <math>u=\sin x</math> gives <math>du=\cos x\,dx</math> and the limits of integration become <math>u=\sin 0=0</math> and <math>u=\sin(\pi/2)=1</math>. The integral is | |
{{Fristående formel||<math>\int_{0}^{\pi/2} \sin^3 x\,\cos x \, dx = \int_{0}^{1} u^3\,du = \Bigl[\,\tfrac{1}{4}u^4\,\Bigr]_{0}^{1} = \tfrac{1}{4} - 0 = \tfrac{1}{4}\,\mbox{.}</math>}} | {{Fristående formel||<math>\int_{0}^{\pi/2} \sin^3 x\,\cos x \, dx = \int_{0}^{1} u^3\,du = \Bigl[\,\tfrac{1}{4}u^4\,\Bigr]_{0}^{1} = \tfrac{1}{4} - 0 = \tfrac{1}{4}\,\mbox{.}</math>}} | ||
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<center>{{:2.2 - Figur - Area under y = sin³x cos x resp. y = u³}}</center> | <center>{{:2.2 - Figur - Area under y = sin³x cos x resp. y = u³}}</center> | ||
{| width="80%" align="center" | {| width="80%" align="center" | ||
- | ||<small> | + | ||<small> The figure on the left shows the graph of the integrand sin³''x'' cos ''x'' and the figure on the right the graph of integrand ''u''³ which is obtained after the variable substitution. The change of variable modifies the integrand and the interval of the integration. The integrals value, the size of the area, is not changed however. </small> |
|} | |} | ||
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<div class="exempel"> | <div class="exempel"> | ||
- | ''' | + | ''' Example 6''' |
- | + | Examine the following calculation | |
{{Fristående formel||<math>\int_{-\pi/2}^{\pi/2} \frac{\cos x}{\sin^2 x}\, dx = \left[\,\begin{align*} &u = \sin x\\ &du = \cos x \, dx\\ &u(-\pi/2) = -1\\ &u (\pi/2) = 1\end{align*}\,\right ] = \int_{-1}^{1} \frac{1}{u^2} \, du = \Bigl[\, -\frac{1}{u}\, \Bigr]_{-1}^{1} = -1 - 1 = -2\,\mbox{.}</math>}} | {{Fristående formel||<math>\int_{-\pi/2}^{\pi/2} \frac{\cos x}{\sin^2 x}\, dx = \left[\,\begin{align*} &u = \sin x\\ &du = \cos x \, dx\\ &u(-\pi/2) = -1\\ &u (\pi/2) = 1\end{align*}\,\right ] = \int_{-1}^{1} \frac{1}{u^2} \, du = \Bigl[\, -\frac{1}{u}\, \Bigr]_{-1}^{1} = -1 - 1 = -2\,\mbox{.}</math>}} | ||
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- | + | This calculation, however, is wrong, which is due to the fact that <math>f(u)=1/u^2</math> is not continuous '''throughout''' the interval <math>[-1,1]</math>. | |
- | + | A necessary condition in the theory is that <math>f(u(x))</math> be defined and continuous for all values which <math>u(x)</math> can take in the interval in question. Otherwise one cannot be certain that the substitution <math>u=u(x)</math> will work. | |
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Version vom 09:36, 22. Jul. 2008
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Content:
- Integration by substitution
Learning outcomes:
After this section, you will have learned to:
- Understand the derivation of the formula for variable substitution .
- Solve easier integration problems that require rewriting and / or substitution in one of the steps.
- Know how the limits of integration are to be changed after a variable substitution.
- Know when variable substitution is allowed.
Variable substitution
When you cannot directly determine a primitive function by using the usual rules of differentiation ”in the opposite direction”, method, other methods or techniques are needed . One such is variable substitution, which can be said to be based on the rule for the differentiation of composite functions — the so-called chain rule.
The chain rule \displaystyle \ \frac{d}{dx}f(u(x)) = f^{\,\prime} (u(x)) \cdot u'(x)\ can be written in integral form as
- REDIRECT Template:Abgesetzte Formel
or,
- REDIRECT Template:Abgesetzte Formel
where F is a primitive function of f. We compare this with the formula
- REDIRECT Template:Abgesetzte Formel
We can see that we have replaced the term \displaystyle u(x) with variable \displaystyle u and the \displaystyle u'(x)\, dx with \displaystyle du. One thus can transform the more complicated integrand \displaystyle f(u(x)) \cdot u'(x) (with \displaystyle x as the variable) to the, let us hope, easier \displaystyle f(u) (with the \displaystyle u as the variable). The method is called variable substitution and can be used when the integrand can be written in the form \displaystyle f(u(x)) \cdot u'(x).
Note 1 The method is based on the assumption that all the conditions for integration are satisfied; that is \displaystyle u(x) is differentiable in the interval in question, and that \displaystyle f is continuous for all values of \displaystyle u in the range, that is for all the values that \displaystyle u can take on in the interval.
Note 2 Replacing \displaystyle u'(x) \, dx with \displaystyle du also may be justified by studying the transition from the increment ratio to the derivative:
- REDIRECT Template:Abgesetzte Formel
which, as \displaystyle \Delta x goes towards zero can be considered as a formal transition between variables
- REDIRECT Template:Abgesetzte Formel
ie., a small change, \displaystyle dx, in the variable \displaystyle x gives rise to an approximate change \displaystyle u'(x)\,dx in the variable \displaystyle u.
Example 1
Determine the integral\displaystyle \ \int 2 x\, e^{x^2} \, dx.
If one puts \displaystyle u(x)= x^2, one gets \displaystyle u'(x)= 2x. The variable substitution replaces \displaystyle e^{x^2} with \displaystyle e^u and \displaystyle u'(x)\,dx, i.e. \displaystyle 2x\,dx, with \displaystyle du
- REDIRECT Template:Abgesetzte Formel
Example 2
Determine the integral \displaystyle \ \int (x^3 + 1)^3 \cdot x^2 \, dx.
Put \displaystyle u=x^3 + 1. This means \displaystyle u'=3x^2, or \displaystyle du= 3x^2\, dx, and
- REDIRECT Template:Abgesetzte Formel
Example 3
Determine the integral \displaystyle \ \int \tan x \, dx\,\mbox{,}\ \ where \displaystyle -\pi/2 < x < \pi/2.
After rewriting \displaystyle \tan x as \displaystyle \sin x/\cos x we substitute \displaystyle u=\cos x,
- REDIRECT Template:Abgesetzte Formel
The limits of integration during variable substitution.
When calculating definite integrals, such as an area, one can go about using variable substitution in two ways. Either one can calculate the integral as usual and then switch back to the original variable and insert the original limits of integration. Alternatively one can change the limits of integration simultaneously with the variable substitution. The two methods are illustrated in the following example.
Example 4
Determine the integral \displaystyle \ \int_{0}^{2} \frac{e^x}{1 + e^x} \, dx.
Method 1
Put \displaystyle u=e^x which gives that \displaystyle u'= e^x and \displaystyle du= e^x\,dx
- REDIRECT Template:Abgesetzte Formel
Note that the limits of integration must be written in the form \displaystyle x = 0 and \displaystyle x = 2 when the variable of integration is not \displaystyle x. it is wrong to write
- REDIRECT Template:Abgesetzte Formel
Method 2
Put \displaystyle u=e^x which gives that \displaystyle u'= e^x and \displaystyle du= e^x\, dx. The limit of integration \displaystyle x=0 is equivalent to \displaystyle u=e^0 = 1 and \displaystyle x=2 is equivalent to \displaystyle u=e^2
- REDIRECT Template:Abgesetzte Formel
Example 5
Determine the integral \displaystyle \ \int_{0}^{\pi/2} \sin^3 x\,\cos x \, dx.
The substitution \displaystyle u=\sin x gives \displaystyle du=\cos x\,dx and the limits of integration become \displaystyle u=\sin 0=0 and \displaystyle u=\sin(\pi/2)=1. The integral is
- REDIRECT Template:Abgesetzte Formel
The figure on the left shows the graph of the integrand sin³x cos x and the figure on the right the graph of integrand u³ which is obtained after the variable substitution. The change of variable modifies the integrand and the interval of the integration. The integrals value, the size of the area, is not changed however. |
Example 6
Examine the following calculation
- REDIRECT Template:Abgesetzte Formel
This calculation, however, is wrong, which is due to the fact that \displaystyle f(u)=1/u^2 is not continuous throughout the interval \displaystyle [-1,1]. A necessary condition in the theory is that \displaystyle f(u(x)) be defined and continuous for all values which \displaystyle u(x) can take in the interval in question. Otherwise one cannot be certain that the substitution \displaystyle u=u(x) will work. |
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