Lösung 2.1:2b

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K (Lösning 2.1:2b moved to Solution 2.1:2b: Robot: moved page)
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There is no ready made standard formula for a primitive function to our integrand, but if we expand
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<center> [[Image:2_1_2b.gif]] </center>
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<math>\begin{align}
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& \int\limits_{-1}^{2}{\left( x-2 \right)\left( x+1 \right)\,dx}=\int\limits_{-1}^{2}{\left( x^{2}+x-2x-2 \right)}\,dx \\
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& \int\limits_{-1}^{2}{\left( x^{2}-x-2 \right)}\,dx \\
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\end{align}</math>
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and write the last integral as
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<math>\int\limits_{-1}^{2}{\left( x^{2}-x^{1}-2x^{0} \right)}\,dx</math>
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we see that the integrand consists of three terms of the type
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<math>x^{n}</math>
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and we can directly write down a primitive function:
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<math>\begin{align}
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& \int\limits_{-1}^{2}{\left( x^{2}-x^{1}-2x^{0} \right)}\,dx=\left[ \frac{x^{3}}{3}-\frac{x^{2}}{2}-2\centerdot \frac{x}{1} \right]_{-1}^{2} \\
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& =\frac{2^{3}}{3}-\frac{2^{2}}{2}-2\centerdot \frac{2}{1}-\left( \frac{\left( -1 \right)^{3}}{3}-\frac{\left( -1 \right)^{2}}{2}-2\centerdot \frac{\left( -1 \right)}{1} \right) \\
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& =\frac{8}{3}-\frac{4}{2}-4-\left( -\frac{1}{3}-\frac{1}{2}+2 \right) \\
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& =\frac{16-12-24+2+3-12}{6}=-\frac{27}{6}=-\frac{9}{2} \\
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\end{align}</math>

Version vom 12:57, 17. Okt. 2008

There is no ready made standard formula for a primitive function to our integrand, but if we expand


\displaystyle \begin{align} & \int\limits_{-1}^{2}{\left( x-2 \right)\left( x+1 \right)\,dx}=\int\limits_{-1}^{2}{\left( x^{2}+x-2x-2 \right)}\,dx \\ & \int\limits_{-1}^{2}{\left( x^{2}-x-2 \right)}\,dx \\ \end{align}

and write the last integral as


\displaystyle \int\limits_{-1}^{2}{\left( x^{2}-x^{1}-2x^{0} \right)}\,dx

we see that the integrand consists of three terms of the type \displaystyle x^{n} and we can directly write down a primitive function:


\displaystyle \begin{align} & \int\limits_{-1}^{2}{\left( x^{2}-x^{1}-2x^{0} \right)}\,dx=\left[ \frac{x^{3}}{3}-\frac{x^{2}}{2}-2\centerdot \frac{x}{1} \right]_{-1}^{2} \\ & =\frac{2^{3}}{3}-\frac{2^{2}}{2}-2\centerdot \frac{2}{1}-\left( \frac{\left( -1 \right)^{3}}{3}-\frac{\left( -1 \right)^{2}}{2}-2\centerdot \frac{\left( -1 \right)}{1} \right) \\ & =\frac{8}{3}-\frac{4}{2}-4-\left( -\frac{1}{3}-\frac{1}{2}+2 \right) \\ & =\frac{16-12-24+2+3-12}{6}=-\frac{27}{6}=-\frac{9}{2} \\ \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.1:2b“