Lösung 2.1:2b

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
Zeile 1: Zeile 1:
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
-
 
+
{{Displayed math||<math>\begin{align}
-
<math>\begin{align}
+
\int\limits_{-1}^{2} (x-2)(x+1)\,dx
-
& \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} (x^2+x-2x-2)\,dx\\[5pt]
-
& \int\limits_{-1}^{2}{\left( x^{2}-x-2 \right)}\,dx \\
+
&= \int\limits_{-1}^{2} (x^2-x-2)\,dx
-
\end{align}</math>
+
\end{align}</math>}}
and write the last integral as
and write the last integral as
-
 
+
{{Displayed math||<math>\int\limits_{-1}^{2} (x^2-x^1-2x^0)\,dx</math>}}
-
<math>\int\limits_{-1}^{2}{\left( x^{2}-x^{1}-2x^{0} \right)}\,dx</math>
+
-
we see that the integrand consists of three terms of the type
+
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,
-
<math>x^{n}</math>
+
-
and we can directly write down a primitive function:
+
-
 
+
-
<math>\begin{align}
+
{{Displayed math||<math>\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} \\
+
\int\limits_{-1}^{2} (x^2-x^1-2x^0)\,dx
-
& =\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) \\
+
&= \Bigl[\ \frac{x^3}{3} - \frac{x^2}{2} - 2\cdot\frac{x}{1}\ \Bigr]_{-1}^{2}\\[5pt]
-
& =\frac{8}{3}-\frac{4}{2}-4-\left( -\frac{1}{3}-\frac{1}{2}+2 \right) \\
+
&= \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{16-12-24+2+3-12}{6}=-\frac{27}{6}=-\frac{9}{2} \\
+
&= \frac{8}{3} - \frac{4}{2} - 4 - \Bigl(-\frac{1}{3}-\frac{1}{2}+2\Bigr)\\[5pt]
-
\end{align}</math>
+
&= \frac{16-12-24+2+3-12}{6}\\[5pt]
 +
&= -\frac{27}{6}\\[5pt]
 +
&= -\frac{9}{2}\,\textrm{.}
 +
\end{align}</math>}}

Version vom 12:45, 21. 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} (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“