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Lösung 2.1:2b

Aus Online Mathematik Brückenkurs 2

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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


2−1x−2x+1dx=2−1x2+x−2x−2dx2−1x2−x−2dx

and write the last integral as


2−1x2−x1−2x0dx 

we see that the integrand consists of three terms of the type xn and we can directly write down a primitive function:


2−1x2−x1−2x0dx=3x3−2x2−2x12−1=323−222−212−3−13−2−12−21−1=38−24−4−−31−21+2=616−12−24+2+3−12=−627=−29

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