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Lösung 2.1:2b

Aus Online Mathematik Brückenkurs 2

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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
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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

2−1(x−2)(x+1)dx=2−1(x2+x−2x−2)dx=2−1(x2−x−2)dx

and write the last integral as

2−1(x2−x1−2x0)dx 

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

2−1(x2−x1−2x0)dx= 3x3−2x2−2x1 2−1=323−222−212−3(−1)3−2(−1)2−21(−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“