2.2 Integration durch Substitution
Aus Online Mathematik Brückenkurs 2
| Theorie | Übungen |
Inhalt:
- Integration durch Substitution
Lernziele:
Nach diesem Abschnitt sollten Sie folgendes können:
- Die Herleitung der Formel für die Integration durch Substitution verstehen.
- Integrale mit Integration durch Substitution lösen.
- Know how the limits of integration are to be changed after a variable substitution.
- Know when substitution is allowed.
Integration durch Substitution
Wenn man eine Funktion nicht direkt integrieren kann, kann man die Funktion manchmal durch eine Substitution integrieren. Die Formel für die Integration durch Substitution ist einfach die Kettenregel für Ableitungen rückwärts.
Die Kettenregel \displaystyle \ \frac{d}{dx}f(u(x)) = f^{\,\prime} (u(x)) \, u'(x)\ kann auf Integralform geschrieben werden:
| \displaystyle \int f^{\,\prime}(u(x)) \, u'(x) \, dx = f(u(x)) + C |
oder,
| \displaystyle \int f(u(x)) \, u'(x) \, dx = F (u(x)) + C\,\mbox{,} |
Wo F eine Stammfunktion von f ist. Wir vergleichen diese Frmel mit der normalen Intagrationsformel
| \displaystyle \int f(u) \, du = F(u) + C\,\mbox{.} |
und sehen dass wir die Variable \displaystyle u(x) mit der Variable \displaystyle u ersetzt haben, und den Term \displaystyle u'(x)\, dx mit \displaystyle du. Daher kann man den komplizierteren Integranden \displaystyle f(u(x)) \, u'(x) ersetzen (mit \displaystyle x als Variable) mit den einfacheren Ausdruck \displaystyle f(u) (mit \displaystyle u als Variable). Dies wird Substitution genannt, und kann verwendet werden wenn der Integrand auf der Form \displaystyle f(u(x)) \, u'(x) ist.
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:
| \displaystyle \lim_{\Delta x \to 0} \frac{\Delta u}{\Delta x} = \frac{du}{dx} = u'(x)\,\mbox{,} |
which, as \displaystyle \Delta x goes towards zero can be considered as a formal transition between variables
| \displaystyle \Delta u \approx u'(x) \Delta x \quad \to \quad du = u'(x) \, dx\,\mbox{,} |
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.
Beispiel 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
| \displaystyle \int 2 x\,e^{x^2} \, dx = \int e^{x^2} \times 2x \, dx = \int e^u \, du = e^u + C = e^{x^2} + C\,\mbox{.} |
Beispiel 2
Determine the integral \displaystyle \ \int (x^3 + 1)^3 \, x^2 \, dx.
Put \displaystyle u=x^3 + 1. This means \displaystyle u'=3x^2, or \displaystyle du= 3x^2\, dx, and
| \displaystyle \begin{align*}\int (x^3 + 1)^3 x^2 \, dx &= \int \frac{ (x^3 + 1)^3}{3} \times 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*} |
Beispiel 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,
| \displaystyle \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*} |
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.
Beispiel 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
| \displaystyle \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*} |
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
| \displaystyle \int_{0}^{2} \frac{e^x}{1 + e^x} \, dx = \int_{0}^{2} \frac{1}{1 + u} \, du \quad \text{ etc.} |
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
| \displaystyle \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{.} |
Beispiel 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
| \displaystyle \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{.} |
![[Image]](/wikis/2009/bridgecourse2-TU-Berlin/img_auth.php/metapost/6/9/c/69c86b3fbe07b4e56291d13e076b36bf.png)
| 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. |
Beispiel 6
Examine the following calculation
| \displaystyle \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{.} |
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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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![[Image]](/wikis/2009/bridgecourse2-TU-Berlin/img_auth.php/metapost/8/b/a/8babf53acdc64bf07e2b6c6b336fa9fd.png)
