Lösung 2.3:2a

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Had the integral instead been
Had the integral instead been
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{{Displayed math||<math>\int e^{\sqrt{x}}\cdot\frac{1}{2\sqrt{x}}\,dx</math>}}
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<math>\int{e^{\sqrt{x}}}\centerdot \frac{1}{2\sqrt{x}}\,dx</math>
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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,
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it is quite obvious that we would substitute
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{{Displayed math||<math>\begin{align}
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<math>u=\sqrt{x}</math>, but we are missing a factor
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\int e^{\sqrt{x}}\,dx
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<math>\frac{1}{2\sqrt{x}}</math>
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&= \int e^{\sqrt{x}}\cdot 2\sqrt{x}\cdot \frac{1}{2\sqrt{x}}\,dx\\[5pt]
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which would take account of the derivative of
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&= \left\{\begin{align}
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<math>u</math>
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u &= \sqrt{x}\\[5pt]
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which is needed when
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du &= \bigl(\sqrt{x}\,\bigr)'\,dx = \dfrac{1}{2\sqrt{x}}\,dx
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<math>dx</math>
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\end{align}\right\}\\[5pt]
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is replaced by
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&= \int e^{u}\cdot 2u\,du\,\textrm{.}
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<math>du</math>. In spite of this, we can try the substitution
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\end{align}</math>}}
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<math>u=\sqrt{x}</math>
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if we multiply top and bottom by what is missing:
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 +
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),
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<math>\begin{align}
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{{Displayed math||<math>\begin{align}
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& \int{e^{\sqrt{x}}\,dx=\int{e^{\sqrt{x}}\centerdot }}2\sqrt{x}\centerdot \frac{1}{2\sqrt{x}}\,dx \\
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\int e^u\cdot 2u\,du
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& =\left\{ \begin{matrix}
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&= e^u\cdot 2u - \int e^u\cdot 2\,du\\[5pt]
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u=\sqrt{x} \\
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&= 2ue^u - 2\int e^u\,du\\[5pt]
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du=\left( \sqrt{x} \right)^{\prime }\,dx=\frac{1}{2\sqrt{x}}\,dx \\
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&= 2ue^u - 2e^u + C\\[5pt]
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\end{matrix} \right\} \\
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&= 2(u-1)e^u + C\,\textrm{.}
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& =\int{e^{u}\centerdot 2u\,du} \\
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\end{align}</math>}}
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\end{align}</math>
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 +
If we substitute back <math>u=\sqrt{x}</math>, we obtain the answer
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Now, we obtain instead another, not entirely simple, integral, but we can calculate the new integral by partial integration (
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{{Displayed math||<math>\int e^{\sqrt{x}}\,dx = 2(\sqrt{x}-1)e^{\sqrt{x}} + C\,\textrm{.}</math>}}
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<math>\text{2}u</math>
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is the factor that we differentiate and
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<math>e^{u}</math>
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is the factor that we integrate),
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<math>\begin{align}
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& \int{e^{u}\centerdot 2u\,du}=e^{u}\centerdot 2u-\int{e^{u}\centerdot 2\,du} \\
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& =2ue^{u}-2\int{e^{u}\,du} \\
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& =2ue^{u}-2e^{u}+C \\
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& =2\left( u-1 \right)e^{u}+C \\
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\end{align}</math>
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If we substitute back
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<math>u=\sqrt{x}</math>, we obtain the answer
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<math>\int{e^{\sqrt{x}}\,dx=2\left( \sqrt{x}-1 \right)}e^{\sqrt{x}}+C</math>
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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 08:37, 29. Okt. 2008

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.

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