Lösung 2.3:1a

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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The formula for integration by parts reads
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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{{Abgesetzte Formel||<math>\int f(x)g(x)\,dx = F(x)g(x) - \int F(x)g'(x)\,dx\,,</math>}}
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>.
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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In the integral
In the integral
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{{Displayed math||<math>\int 2xe^{-x}\,dx\,,</math>}}
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{{Abgesetzte Formel||<math>\int 2xe^{-x}\,dx\,,</math>}}
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,
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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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\int 2x\cdot e^{-x}\,dx
\int 2x\cdot e^{-x}\,dx
&= 2x\cdot \bigl(-e^{-x}\bigr) - \int 2\cdot \bigl(-e^{-x}\bigr)\,dx\\[5pt]
&= 2x\cdot \bigl(-e^{-x}\bigr) - \int 2\cdot \bigl(-e^{-x}\bigr)\,dx\\[5pt]
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It remains only to integrate <math>e^{-x}</math> and we are finished,
It remains only to integrate <math>e^{-x}</math> and we are finished,
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\phantom{\int 2x\cdot e^{-x}\,dx}{}
\phantom{\int 2x\cdot e^{-x}\,dx}{}
&= \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]
&= \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]

Version vom 13:03, 10. Mär. 2009

The formula for integration by parts reads

\displaystyle \int f(x)g(x)\,dx = F(x)g(x) - \int F(x)g'(x)\,dx\,,

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

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

In the integral

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

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

\displaystyle \begin{align}

\int 2x\cdot e^{-x}\,dx &= 2x\cdot \bigl(-e^{-x}\bigr) - \int 2\cdot \bigl(-e^{-x}\bigr)\,dx\\[5pt] &= -2xe^{-x} + 2\int e^{-x}\,dx\,\textrm{.} \end{align}

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

\displaystyle \begin{align}

\phantom{\int 2x\cdot e^{-x}\,dx}{} &= \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] &= -2xe^{-x} - 2e^{-x} + C\\[5pt] &= -2(x+1)e^{-x} + C\,\textrm{.} \end{align}

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