Lösung 2.2:2c

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If we focus on the integrand, then the substitution
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<center> [[Image:2_2_2c.gif]] </center>
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<math>u=\text{3}x+\text{1}</math>
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seems suitable, since we then get
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<math>\sqrt{u}</math>
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which we can integrate. There is also no risk involved in using a linear substitution such as
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<math>u=\text{3}x+\text{1}</math>, because the relation between
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<math>dx\text{ }</math>
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and
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<math>du</math>
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will be a constant factor,
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<math>du=\left( \text{3}x+\text{1} \right)^{1}\,dx=3\,dx</math>
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which does not cause any problems.
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We obtain
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<math>\begin{align}
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& \int\limits_{0}^{5}{\sqrt{3x+1}}\,dx=\left\{ \begin{matrix}
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u=\text{3}x+\text{1} \\
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du=3\,dx \\
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\end{matrix} \right\}=\frac{1}{3}\int\limits_{1}^{16}{\sqrt{u}}\,du \\
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& =\frac{1}{3}\int\limits_{1}^{16}{u^{{1}/{2}\;}}\,du=\frac{1}{3}\left[ \frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1} \right]_{1}^{16} \\
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& =\frac{1}{3}\left[ \frac{2}{3}u\sqrt{u} \right]_{1}^{16}=\frac{2}{9}\left( 16\sqrt{16}-1\sqrt{1} \right) \\
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& =\frac{2}{9}\left( 16\centerdot 4-1 \right)=\frac{2\centerdot 63}{9}=14 \\
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\end{align}</math>

Version vom 12:10, 20. Okt. 2008

If we focus on the integrand, then the substitution \displaystyle u=\text{3}x+\text{1} seems suitable, since we then get \displaystyle \sqrt{u} which we can integrate. There is also no risk involved in using a linear substitution such as \displaystyle u=\text{3}x+\text{1}, because the relation between \displaystyle dx\text{ } and \displaystyle du will be a constant factor,


\displaystyle du=\left( \text{3}x+\text{1} \right)^{1}\,dx=3\,dx


which does not cause any problems.

We obtain


\displaystyle \begin{align} & \int\limits_{0}^{5}{\sqrt{3x+1}}\,dx=\left\{ \begin{matrix} u=\text{3}x+\text{1} \\ du=3\,dx \\ \end{matrix} \right\}=\frac{1}{3}\int\limits_{1}^{16}{\sqrt{u}}\,du \\ & =\frac{1}{3}\int\limits_{1}^{16}{u^{{1}/{2}\;}}\,du=\frac{1}{3}\left[ \frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1} \right]_{1}^{16} \\ & =\frac{1}{3}\left[ \frac{2}{3}u\sqrt{u} \right]_{1}^{16}=\frac{2}{9}\left( 16\sqrt{16}-1\sqrt{1} \right) \\ & =\frac{2}{9}\left( 16\centerdot 4-1 \right)=\frac{2\centerdot 63}{9}=14 \\ \end{align}

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