Loading http://wiki.math.se/jsMath/fonts/msam10/def.js

Aus Online Mathematik Brückenkurs 2

Version vom 14:48, 16. Sep. 2008 (bearbeiten) Tekbot (Diskussion | Beiträge) K (Lösning 2.3:2b moved to Solution 2.3:2b: Robot: moved page) ← Zum vorherigen Versionsunterschied		Version vom 13:50, 22. Okt. 2008 (bearbeiten) (rückgängig) Ian (Diskussion | Beiträge) Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
-	{{NAVCONTENT_START}}	+	We have a product of two factors in the integrand, so a partial integration does not seem unreasonable. There is nevertheless a problem as regards which factor should be differentiated and which should be integrated. If we choose to differentiate
-	<center> [[Image:2_3_2b-1(2).gif]] </center>	+	<math>x^{\text{3}}</math>
-	{{NAVCONTENT_STOP}}	+	(so as to reduce its exponent by
-	{{NAVCONTENT_START}}	+	<math>\text{1}</math>), we need to find a primitive function for
-	<center> [[Image:2_3_2b-2(2).gif]] </center>	+	<math>e^{x^{2}}</math>, and how do we do that? If, on the other hand, we integrate
-	{{NAVCONTENT_STOP}}	+	<math>x^{\text{3}}</math>
		+	and differentiate
		+	<math>e^{x^{2}}</math>, we get
		+
		+
		+	<math>\begin{align}
		+	& \int{x^{3}e^{x^{2}}\,dx=\frac{x^{4}}{4}}\centerdot e^{x^{2}}-\int{\frac{x^{4}}{4}}\centerdot e^{x^{2}}2x\,dx \\
		+	& =\frac{1}{4}x^{4}e^{x^{2}}-\frac{1}{2}\int{x^{5}e^{x^{2}}\,dx} \\
		+	\end{align}</math>
		+
		+	which just seems to make the integral harder. The solution is instead to substitute
		+	<math>u=x^{2}</math>. If we write the integral as
		+
		+
		+	<math>\int\limits_{0}^{1}{x^{3}e^{x^{2}}\,dx}=\int\limits_{0}^{1}{x^{2}e^{x^{2}}x\,dx}</math>
		+
		+
		+	we see that the expression
		+	<math>''x\,dx''</math>
		+	can be replaced by
		+	<math>du</math>
		+	and the rest of the integrand contains only
		+	<math>x</math>
		+	in the form of
		+	<math>x^{\text{2}}</math>. The substitution gives
		+
		+
		+	<math>\begin{align}
		+	& \int\limits_{0}^{1}{x^{3}e^{x^{2}}\,dx}=\int\limits_{0}^{1}{x^{2}e^{x^{2}}x\,dx}=\left\{ \begin{matrix}
		+	u=x^{2} \\
		+	du=\left( x^{2} \right)^{\prime }\,dx=2x\,dx \\
		+	\end{matrix} \right\} \\
		+	& =\int\limits_{0}^{1}{ue^{u}\frac{1}{2}\,du}=\frac{1}{2}\int\limits_{0}^{1}{ue^{u}\,du} \\
		+	\end{align}</math>
		+
		+
		+	We can then calculate this integral be partial integration, where we differentiate away the factor
		+	<math>u</math>:
		+
		+
		+	<math>\begin{align}
		+	& \frac{1}{2}\int\limits_{0}^{1}{ue^{u}\,du}=\frac{1}{2}\left[ ue^{u} \right]_{0}^{1}-\frac{1}{2}\int\limits_{0}^{1}{1\centerdot e^{u}\,du} \\
		+	& =\frac{1}{2}\left( 1\centerdot e^{1}-0 \right)-\frac{1}{2}\left[ e^{u} \right]_{0}^{1} \\
		+	& =\frac{1}{2}e-\frac{1}{2}\left( e^{1}-e^{0} \right) \\
		+	& =\frac{1}{2}e-\frac{1}{2}e+\frac{1}{2}=\frac{1}{2} \\
		+	\end{align}</math>

---

Version vom 13:50, 22. Okt. 2008

We have a product of two factors in the integrand, so a partial integration does not seem unreasonable. There is nevertheless a problem as regards which factor should be differentiated and which should be integrated. If we choose to differentiate x3 (so as to reduce its exponent by 1), we need to find a primitive function for ex2, and how do we do that? If, on the other hand, we integrate x3 and differentiate ex2, we get

\displaystyle \begin{align} & \int{x^{3}e^{x^{2}}\,dx=\frac{x^{4}}{4}}\centerdot e^{x^{2}}-\int{\frac{x^{4}}{4}}\centerdot e^{x^{2}}2x\,dx \\ & =\frac{1}{4}x^{4}e^{x^{2}}-\frac{1}{2}\int{x^{5}e^{x^{2}}\,dx} \\ \end{align}

which just seems to make the integral harder. The solution is instead to substitute \displaystyle u=x^{2}. If we write the integral as

\displaystyle \int\limits_{0}^{1}{x^{3}e^{x^{2}}\,dx}=\int\limits_{0}^{1}{x^{2}e^{x^{2}}x\,dx}

we see that the expression \displaystyle ''x\,dx'' can be replaced by \displaystyle du and the rest of the integrand contains only \displaystyle x in the form of \displaystyle x^{\text{2}}. The substitution gives

\displaystyle \begin{align} & \int\limits_{0}^{1}{x^{3}e^{x^{2}}\,dx}=\int\limits_{0}^{1}{x^{2}e^{x^{2}}x\,dx}=\left\{ \begin{matrix} u=x^{2} \\ du=\left( x^{2} \right)^{\prime }\,dx=2x\,dx \\ \end{matrix} \right\} \\ & =\int\limits_{0}^{1}{ue^{u}\frac{1}{2}\,du}=\frac{1}{2}\int\limits_{0}^{1}{ue^{u}\,du} \\ \end{align}

We can then calculate this integral be partial integration, where we differentiate away the factor \displaystyle u:

\displaystyle \begin{align} & \frac{1}{2}\int\limits_{0}^{1}{ue^{u}\,du}=\frac{1}{2}\left[ ue^{u} \right]_{0}^{1}-\frac{1}{2}\int\limits_{0}^{1}{1\centerdot e^{u}\,du} \\ & =\frac{1}{2}\left( 1\centerdot e^{1}-0 \right)-\frac{1}{2}\left[ e^{u} \right]_{0}^{1} \\ & =\frac{1}{2}e-\frac{1}{2}\left( e^{1}-e^{0} \right) \\ & =\frac{1}{2}e-\frac{1}{2}e+\frac{1}{2}=\frac{1}{2} \\ \end{align}