Lösung 2.2:2c

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If we focus on the integrand, then the substitution
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If we focus on the integrand, then the substitution <math>u=3x+1</math> seems suitable, since we then get <math>\sqrt{u}</math> which we can integrate. There is also no risk involved in using a linear substitution such as <math>u=3x+1</math>, because the relation between <math>dx</math> and <math>du</math> will be a constant factor,
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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)^{\prime }\,dx=3\,dx</math>
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{{Displayed math||<math>du = (3x+1)'\,dx = 3\,dx\,,</math>}}
which does not cause any problems.
which does not cause any problems.
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We obtain
We obtain
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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\int\limits_0^5 \sqrt{3x+1}\,dx
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& \int\limits_{0}^{5}{\sqrt{3x+1}}\,dx=\left\{ \begin{matrix}
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&= \left\{\begin{align}
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u=\text{3}x+\text{1} \\
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u &= 3x+1\\[5pt]
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du=3\,dx \\
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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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\end{align}\right\} = \frac{1}{3}\int\limits_1^{16} \sqrt{u}\,du\\[5pt]
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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}\int\limits_1^{16} u^{1/2}\,du = \frac{1}{3}\biggl[\ \frac{u^{1/2+1}}{\tfrac{1}{2}+1}\ \biggr]_1^{16}\\[5pt]
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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{1}{3}\Bigl[\ \frac{2}{3}u\sqrt{u}\ \Bigr]_1^{16} = \frac{2}{9}\bigl( 16\sqrt{16}-1\sqrt{1} \bigr)\\[8pt]
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& =\frac{2}{9}\left( 16\centerdot 4-1 \right)=\frac{2\centerdot 63}{9}=14 \\
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&= \frac{2}{9}\bigl( 16\cdot 4-1 \bigr) = \frac{2\cdot 63}{9} = 14\,\textrm{.}
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\end{align}</math>
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\end{align}</math>}}

Version vom 13:36, 28. Okt. 2008

If we focus on the integrand, then the substitution \displaystyle u=3x+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=3x+1, because the relation between \displaystyle dx and \displaystyle du will be a constant factor,

\displaystyle du = (3x+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{align} u &= 3x+1\\[5pt] du &= 3\,dx \end{align}\right\} = \frac{1}{3}\int\limits_1^{16} \sqrt{u}\,du\\[5pt] &= \frac{1}{3}\int\limits_1^{16} u^{1/2}\,du = \frac{1}{3}\biggl[\ \frac{u^{1/2+1}}{\tfrac{1}{2}+1}\ \biggr]_1^{16}\\[5pt] &= \frac{1}{3}\Bigl[\ \frac{2}{3}u\sqrt{u}\ \Bigr]_1^{16} = \frac{2}{9}\bigl( 16\sqrt{16}-1\sqrt{1} \bigr)\\[8pt] &= \frac{2}{9}\bigl( 16\cdot 4-1 \bigr) = \frac{2\cdot 63}{9} = 14\,\textrm{.} \end{align}

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