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Lösung 2.3:1a

Aus Online Mathematik Brückenkurs 2

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K
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>.
Zeile 10: Zeile 10:
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]
Zeile 22: Zeile 22:
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

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“