Lösung 2.3:1a

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The formula for partial integration reads
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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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where
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<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
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<math>{g}'\left( x \right)</math>
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is a derivative of
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<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
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<math>f\left( x \right)</math>
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which we integrate and a factor
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<math>g\left( x \right)</math>
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which we differentiate. It is only when the product
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<math>F\left( x \right){g}'\left( x \right)</math>
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becomes simpler than
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<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
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<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
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<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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It remains only to integrate
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<math>e^{-x}</math>
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and we are finished:
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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


\displaystyle \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},

where \displaystyle F\left( x \right) is a primitive function of \displaystyle f\left( x \right) and \displaystyle {g}'\left( x \right) is a derivative of \displaystyle g\left( x \right).

If we are to use partial integration, the integrand has to be divided up into two factors, a factor \displaystyle f\left( x \right) which we integrate and a factor \displaystyle g\left( x \right) which we differentiate. It is only when the product \displaystyle F\left( x \right){g}'\left( x \right) becomes simpler than \displaystyle f\left( x \right)g\left( x \right) that there is any point in partially integrating.

In the integral


\displaystyle \int{2xe^{-x}}\,dx,

it can seem appropriate to choose \displaystyle f\left( x \right)=e^{-x} and \displaystyle g\left( x \right)=2x, because then \displaystyle {g}'\left( x \right)=2 and we have only \displaystyle F\left( x \right)=-e^{-x} left,


\displaystyle \begin{align} & \int{2xe^{-x}}\,dx=2x\centerdot \left( -e^{-x} \right)-\int{2\centerdot }\left( -e^{-x} \right)\,dx \\ & =-2xe^{-x}+2\int{e^{-x}\,dx} \\ \end{align}


It remains only to integrate \displaystyle e^{-x} and we are finished:


\displaystyle \begin{align} & =-2xe^{-x}+2\left( -e^{-x} \right)+C \\ & =-2xe^{-x}-2e^{-x}+C \\ & =-2\left( x+1 \right)e^{-x}+C \\ \end{align}

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