Lösung 2.1:2b

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There is no ready made standard formula for a primitive function to our integrand, but if we expand
There is no ready made standard formula for a primitive function to our integrand, but if we expand
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\int\limits_{-1}^{2} (x-2)(x+1)\,dx
\int\limits_{-1}^{2} (x-2)(x+1)\,dx
&= \int\limits_{-1}^{2} (x^2+x-2x-2)\,dx\\[5pt]
&= \int\limits_{-1}^{2} (x^2+x-2x-2)\,dx\\[5pt]
Zeile 9: Zeile 9:
and write the last integral as
and write the last integral as
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{{Displayed math||<math>\int\limits_{-1}^{2} (x^2-x^1-2x^0)\,dx</math>}}
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{{Abgesetzte Formel||<math>\int\limits_{-1}^{2} (x^2-x^1-2x^0)\,dx</math>}}
we see that the integrand consists of three terms of the type <math>x^n</math> and we can directly write down a primitive function,
we see that the integrand consists of three terms of the type <math>x^n</math> and we can directly write down a primitive function,
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\int\limits_{-1}^{2} (x^2-x^1-2x^0)\,dx
\int\limits_{-1}^{2} (x^2-x^1-2x^0)\,dx
&= \Bigl[\ \frac{x^3}{3} - \frac{x^2}{2} - 2\cdot\frac{x}{1}\ \Bigr]_{-1}^{2}\\[5pt]
&= \Bigl[\ \frac{x^3}{3} - \frac{x^2}{2} - 2\cdot\frac{x}{1}\ \Bigr]_{-1}^{2}\\[5pt]

Version vom 12:58, 10. Mär. 2009

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} (x-2)(x+1)\,dx &= \int\limits_{-1}^{2} (x^2+x-2x-2)\,dx\\[5pt] &= \int\limits_{-1}^{2} (x^2-x-2)\,dx \end{align}

and write the last integral as

\displaystyle \int\limits_{-1}^{2} (x^2-x^1-2x^0)\,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} (x^2-x^1-2x^0)\,dx &= \Bigl[\ \frac{x^3}{3} - \frac{x^2}{2} - 2\cdot\frac{x}{1}\ \Bigr]_{-1}^{2}\\[5pt] &= \frac{2^3}{3} - \frac{2^2}{2} - 2\cdot\frac{2}{1} - \Bigl(\frac{(-1)^3}{3} - \frac{(-1)^2}{2} - 2\cdot\frac{(-1)}{1}\Bigr)\\[5pt] &= \frac{8}{3} - \frac{4}{2} - 4 - \Bigl(-\frac{1}{3}-\frac{1}{2}+2\Bigr)\\[5pt] &= \frac{16-12-24+2+3-12}{6}\\[5pt] &= -\frac{27}{6}\\[5pt] &= -\frac{9}{2}\,\textrm{.} \end{align}

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