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Lösung 2.2:3f

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
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K (Lösning 2.2:3f moved to Solution 2.2:3f: Robot: moved page)
Zeile 1: Zeile 1:
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{{NAVCONTENT_START}}
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Let's rewrite the integral somewhat:
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<center> [[Image:2_2_3f.gif]] </center>
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{{NAVCONTENT_STOP}}
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<math>2\sin \sqrt{x}\centerdot \frac{1}{2\sqrt{x}}</math>
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 +
 
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Here, we see that the factor on the right,
 +
<math>\frac{1}{2\sqrt{x}}</math>
 +
is the derivative of the expression
 +
<math>\sqrt{x}</math>, which appears in the factor on the left,
 +
<math>2\sin \sqrt{x}</math>
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With the substitution
 +
<math>u=\sqrt{x}</math>, the integrand can therefore be written as
 +
 
 +
 
 +
<math>2\sin u\centerdot {u}'</math>
 +
 
 +
 
 +
and the integral becomes
 +
 
 +
 
 +
<math>\begin{align}
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& \int{\frac{\sin \sqrt{x}}{\sqrt{x}}}\,dx=\left\{ \begin{matrix}
 +
u=\sqrt{x} \\
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du=\left( \sqrt{x} \right)^{\prime }\,dx=\frac{1}{2\sqrt{x}}\,dx \\
 +
\end{matrix}\, \right\} \\
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& =2\int{\sin u\,du} \\
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& =-2\cos u+C \\
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& =-2\cos \sqrt{x}+C \\
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\end{align}</math>

Version vom 15:38, 28. Okt. 2008

Let's rewrite the integral somewhat:


2sinx12x


Here, we see that the factor on the right, 12x is the derivative of the expression x , which appears in the factor on the left, 2sinx  With the substitution u=x , the integrand can therefore be written as


2sinuu


and the integral becomes


xsinxdx=u=xdu=xdx=12xdx=2sinudu=−2cosu+C=−2cosx+C

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