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 →
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-	{{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>

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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 \displaystyle x^{\text{3}} (so as to reduce its exponent by \displaystyle \text{1}), we need to find a primitive function for \displaystyle e^{x^{2}}, and how do we do that? If, on the other hand, we integrate \displaystyle x^{\text{3}} and differentiate \displaystyle e^{x^{2}}, 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}