Lösung 2.3:2a

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Lösning 2.3:2a moved to Solution 2.3:2a: Robot: moved page)
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
Had the integral instead been
-
<center> [[Image:2_3_2a-1(2).gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
 
-
{{NAVCONTENT_START}}
+
<math>\int{e^{\sqrt{x}}}\centerdot \frac{1}{2\sqrt{x}}\,dx</math>
-
<center> [[Image:2_3_2a-2(2).gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
it is quite obvious that we would substitute
 +
<math>u=\sqrt{x}</math>, but we are missing a factor
 +
<math>\frac{1}{2\sqrt{x}}</math>
 +
which would take account of the derivative of
 +
<math>u</math>
 +
which is needed when
 +
<math>dx</math>
 +
is replaced by
 +
<math>du</math>. In spite of this, we can try the substitution
 +
<math>u=\sqrt{x}</math>
 +
if we multiply top and bottom by what is missing:
 +
 
 +
 
 +
<math>\begin{align}
 +
& \int{e^{\sqrt{x}}\,dx=\int{e^{\sqrt{x}}\centerdot }}2\sqrt{x}\centerdot \frac{1}{2\sqrt{x}}\,dx \\
 +
& =\left\{ \begin{matrix}
 +
u=\sqrt{x} \\
 +
du=\left( \sqrt{x} \right)^{\prime }\,dx=\frac{1}{2\sqrt{x}}\,dx \\
 +
\end{matrix} \right\} \\
 +
& =\int{e^{u}\centerdot 2u\,du} \\
 +
\end{align}</math>
 +
 
 +
 
 +
Now, we obtain instead another, not entirely simple, integral, but we can calculate the new integral by partial integration (
 +
<math>\text{2}u</math>
 +
is the factor that we differentiate and
 +
<math>e^{u}</math>
 +
is the factor that we integrate),
 +
 
 +
 
 +
<math>\begin{align}
 +
& \int{e^{u}\centerdot 2u\,du}=e^{u}\centerdot 2u-\int{e^{u}\centerdot 2\,du} \\
 +
& =2ue^{u}-2\int{e^{u}\,du} \\
 +
& =2ue^{u}-2e^{u}+C \\
 +
& =2\left( u-1 \right)e^{u}+C \\
 +
\end{align}</math>
 +
 
 +
 
 +
If we substitute back
 +
<math>u=\sqrt{x}</math>, we obtain the answer
 +
 
 +
 
 +
<math>\int{e^{\sqrt{x}}\,dx=2\left( \sqrt{x}-1 \right)}e^{\sqrt{x}}+C</math>
 +
 
 +
 
 +
As can be seen, it is possible to mix different integration techniques and often we need to experiment with different approaches before we find the right one.

Version vom 13:25, 22. Okt. 2008

Had the integral instead been


\displaystyle \int{e^{\sqrt{x}}}\centerdot \frac{1}{2\sqrt{x}}\,dx

it is quite obvious that we would substitute \displaystyle u=\sqrt{x}, but we are missing a factor \displaystyle \frac{1}{2\sqrt{x}} which would take account of the derivative of \displaystyle u which is needed when \displaystyle dx is replaced by \displaystyle du. In spite of this, we can try the substitution \displaystyle u=\sqrt{x} if we multiply top and bottom by what is missing:


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


Now, we obtain instead another, not entirely simple, integral, but we can calculate the new integral by partial integration ( \displaystyle \text{2}u is the factor that we differentiate and \displaystyle e^{u} is the factor that we integrate),


\displaystyle \begin{align} & \int{e^{u}\centerdot 2u\,du}=e^{u}\centerdot 2u-\int{e^{u}\centerdot 2\,du} \\ & =2ue^{u}-2\int{e^{u}\,du} \\ & =2ue^{u}-2e^{u}+C \\ & =2\left( u-1 \right)e^{u}+C \\ \end{align}


If we substitute back \displaystyle u=\sqrt{x}, we obtain the answer


\displaystyle \int{e^{\sqrt{x}}\,dx=2\left( \sqrt{x}-1 \right)}e^{\sqrt{x}}+C


As can be seen, it is possible to mix different integration techniques and often we need to experiment with different approaches before we find the right one.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.3:2a“