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Lösung 2.3:1a

Aus Online Mathematik Brückenkurs 2

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K (Lösning 2.3:1a moved to Solution 2.3:1a: Robot: moved page)
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The formula for partial integration reads
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<center> [[Image:2_3_1a-1(2).gif]] </center>
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<math>\int{f\left( x \right)}g\left( x \right)\,dx=F\left( x \right)g\left( x \right)-\int{F\left( x \right){g}'\left( x \right)\,dx}</math>,
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<center> [[Image:2_3_1a-2(2).gif]] </center>
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where
 +
<math>F\left( x \right)</math>
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is a primitive function of
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<math>f\left( x \right)</math>
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and
 +
<math>{g}'\left( x \right)</math>
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is a derivative of
 +
<math>g\left( x \right)</math>.
 +
 
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If we are to use partial integration, the integrand has to be divided up into two factors, a factor
 +
<math>f\left( x \right)</math>
 +
which we integrate and a factor
 +
<math>g\left( x \right)</math>
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which we differentiate. It is only when the product
 +
<math>F\left( x \right){g}'\left( x \right)</math>
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becomes simpler than
 +
<math>f\left( x \right)g\left( x \right)</math>
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that there is any point in partially integrating.
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 +
In the integral
 +
 
 +
 
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<math>\int{2xe^{-x}}\,dx</math>,
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 +
it can seem appropriate to choose
 +
<math>f\left( x \right)=e^{-x}</math>
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and
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<math>g\left( x \right)=2x</math>, because then
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<math>{g}'\left( x \right)=2</math>
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and we have only
 +
<math>F\left( x \right)=-e^{-x}</math>
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left,
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<math>\begin{align}
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& \int{2xe^{-x}}\,dx=2x\centerdot \left( -e^{-x} \right)-\int{2\centerdot }\left( -e^{-x} \right)\,dx \\
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& =-2xe^{-x}+2\int{e^{-x}\,dx} \\
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\end{align}</math>
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 +
 
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It remains only to integrate
 +
<math>e^{-x}</math>
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and we are finished:
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 +
 
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<math>\begin{align}
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& =-2xe^{-x}+2\left( -e^{-x} \right)+C \\
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& =-2xe^{-x}-2e^{-x}+C \\
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& =-2\left( x+1 \right)e^{-x}+C \\
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\end{align}</math>

Version vom 13:24, 21. Okt. 2008

The formula for partial integration reads


fxgxdx=Fxgx−Fxgxdx ,

where Fx  is a primitive function of fx  and gx  is a derivative of gx .

If we are to use partial integration, the integrand has to be divided up into two factors, a factor fx  which we integrate and a factor gx  which we differentiate. It is only when the product Fxgx  becomes simpler than fxgx  that there is any point in partially integrating.

In the integral


2xe−xdx ,

it can seem appropriate to choose fx=e−x  and gx=2x , because then gx=2  and we have only Fx=−e−x  left,


2xe−xdx=2x−e−x−2−e−xdx=−2xe−x+2e−xdx 


It remains only to integrate e−x and we are finished:


=−2xe−x+2−e−x+C=−2xe−x−2e−x+C=−2x+1e−x+C

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.3:1a“