Lösung 2.3:2a
Aus Online Mathematik Brückenkurs 2
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Had the integral instead been | Had the integral instead been | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\int e^{\sqrt{x}}\cdot\frac{1}{2\sqrt{x}}\,dx</math>}} |
it is quite obvious that we would substitute <math>u=\sqrt{x}</math>, but we are missing a factor <math>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 | it is quite obvious that we would substitute <math>u=\sqrt{x}</math>, but we are missing a factor <math>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>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, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\int e^{\sqrt{x}}\,dx | \int e^{\sqrt{x}}\,dx | ||
&= \int e^{\sqrt{x}}\cdot 2\sqrt{x}\cdot \frac{1}{2\sqrt{x}}\,dx\\[5pt] | &= \int e^{\sqrt{x}}\cdot 2\sqrt{x}\cdot \frac{1}{2\sqrt{x}}\,dx\\[5pt] | ||
| Zeile 18: | Zeile 18: | ||
Now, we obtain instead another, not entirely simple, integral, but we can calculate the new integral by partial integration (<math>2u</math> is the factor that we differentiate and <math>e^{u}</math> is the factor that we integrate), | Now, we obtain instead another, not entirely simple, integral, but we can calculate the new integral by partial integration (<math>2u</math> is the factor that we differentiate and <math>e^{u}</math> is the factor that we integrate), | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\int e^u\cdot 2u\,du | \int e^u\cdot 2u\,du | ||
&= e^u\cdot 2u - \int e^u\cdot 2\,du\\[5pt] | &= e^u\cdot 2u - \int e^u\cdot 2\,du\\[5pt] | ||
| Zeile 28: | Zeile 28: | ||
If we substitute back <math>u=\sqrt{x}</math>, we obtain the answer | If we substitute back <math>u=\sqrt{x}</math>, we obtain the answer | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\int e^{\sqrt{x}}\,dx = 2(\sqrt{x}-1)e^{\sqrt{x}} + C\,\textrm{.}</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. | 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 13:04, 10. Mär. 2009
Had the integral instead been
| \displaystyle \int e^{\sqrt{x}}\cdot\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 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}}\cdot 2\sqrt{x}\cdot \frac{1}{2\sqrt{x}}\,dx\\[5pt] &= \left\{\begin{align} u &= \sqrt{x}\\[5pt] du &= \bigl(\sqrt{x}\,\bigr)'\,dx = \dfrac{1}{2\sqrt{x}}\,dx \end{align}\right\}\\[5pt] &= \int e^{u}\cdot 2u\,du\,\textrm{.} \end{align} |
Now, we obtain instead another, not entirely simple, integral, but we can calculate the new integral by partial integration (\displaystyle 2u is the factor that we differentiate and \displaystyle e^{u} is the factor that we integrate),
| \displaystyle \begin{align}
\int e^u\cdot 2u\,du &= e^u\cdot 2u - \int e^u\cdot 2\,du\\[5pt] &= 2ue^u - 2\int e^u\,du\\[5pt] &= 2ue^u - 2e^u + C\\[5pt] &= 2(u-1)e^u + C\,\textrm{.} \end{align} |
If we substitute back \displaystyle u=\sqrt{x}, we obtain the answer
| \displaystyle \int e^{\sqrt{x}}\,dx = 2(\sqrt{x}-1)e^{\sqrt{x}} + C\,\textrm{.} |
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.
