Lösung 2.2:2c
Aus Online Mathematik Brückenkurs 2
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| - | <math>du=\left( \text{3}x+\text{1} \right)^{ | + | <math>du=\left( \text{3}x+\text{1} \right)^{\prime }\,dx=3\,dx</math> |
Version vom 12:39, 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)^{\prime }\,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}
