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Lösung 2.3:1a

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
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The formula for partial integration reads
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The formula for integration by parts reads
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{{Displayed math||<math>\int f(x)g(x)\,dx = F(x)g(x) - \int F(x)g'(x)\,dx\,,</math>}}
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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 <math>F(x)</math> is a primitive function of <math>f(x)</math> and <math>g'(x)</math> is a derivative of <math>g(x)</math>.
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where
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If we are to use integration by parts, the integrand has to be divided up into two factors, a factor <math>f(x)</math> which we integrate and a factor <math>g(x)</math> which we differentiate. It is only when the product <math>F(x)g'(x)</math>
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<math>F\left( x \right)</math>
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becomes simpler than <math>f(x)g(x)</math> that there is any point in integrating by parts.
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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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+
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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
In the integral
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{{Displayed math||<math>\int 2xe^{-x}\,dx\,,</math>}}
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<math>\int{2xe^{-x}}\,dx</math>,
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it can seem appropriate to choose <math>f(x)=e^{-x}</math> and <math>g(x) = 2x</math>, because then <math>g'(x) = 2</math> and we have only <math>F(x) = -e^{-x}</math> left,
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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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{{Displayed math||<math>\begin{align}
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<math>e^{-x}</math>
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\int 2x\cdot e^{-x}\,dx
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and we are finished:
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&= 2x\cdot \bigl(-e^{-x}\bigr) - \int 2\cdot \bigl(-e^{-x}\bigr)\,dx\\[5pt]
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&= -2xe^{-x} + 2\int e^{-x}\,dx\,\textrm{.}
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\end{align}</math>}}
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It remains only to integrate <math>e^{-x}</math> and we are finished,
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<math>\begin{align}
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{{Displayed math||<math>\begin{align}
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& =-2xe^{-x}+2\left( -e^{-x} \right)+C \\
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\phantom{\int 2x\cdot e^{-x}\,dx}{}
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& =-2xe^{-x}-2e^{-x}+C \\
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&= \rlap{-2xe^{-x} + 2\bigl(-e^{-x}\bigr) + C}\phantom{2x\cdot \bigl(-e^{-x}\bigr) - \int 2\cdot \bigl(-e^{-x}\bigr)\,dx}\\[5pt]
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& =-2\left( x+1 \right)e^{-x}+C \\
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&= -2xe^{-x} - 2e^{-x} + C\\[5pt]
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\end{align}</math>
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&= -2(x+1)e^{-x} + C\,\textrm{.}
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\end{align}</math>}}

Version vom 08:08, 29. Okt. 2008

The formula for integration by parts reads

f(x)g(x)dx=F(x)g(x)−F(x)g(x)dx 

where F(x) is a primitive function of f(x) and g(x) is a derivative of g(x).

If we are to use integration by parts, the integrand has to be divided up into two factors, a factor f(x) which we integrate and a factor g(x) which we differentiate. It is only when the product F(x)g(x) becomes simpler than f(x)g(x) that there is any point in integrating by parts.

In the integral

2xe−xdx 

it can seem appropriate to choose f(x)=e−x and g(x)=2x, because then g(x)=2 and we have only F(x)=−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=−2(x+1)e−x+C.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.3:1a“