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Lösung 2.1:4a

Aus Online Mathematik Brückenkurs 2

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K
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 5: Zeile 5:
The area of the region between <math>x=0</math> and <math>x=\pi</math> can therefore be written as
The area of the region between <math>x=0</math> and <math>x=\pi</math> can therefore be written as
-
{{Displayed math||<math>\int\limits_{0}^{\pi} \sin x\,dx</math>}}
+
{{Abgesetzte Formel||<math>\int\limits_{0}^{\pi} \sin x\,dx</math>}}
whilst the area of the remaining region under the ''x''-axis is equal to
whilst the area of the remaining region under the ''x''-axis is equal to
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{{Displayed math||<math>-\int\limits_{\pi}^{5\pi/4} \sin x\,dx</math>}}
+
{{Abgesetzte Formel||<math>-\int\limits_{\pi}^{5\pi/4} \sin x\,dx</math>}}
(note the minus sign in front of the integral).
(note the minus sign in front of the integral).
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The total area becomes
The total area becomes
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
& \int\limits_{0}^{\pi} \sin x\,dx - \int\limits_{\pi}^{5\pi/4} \sin x\,dx\\[5pt]
& \int\limits_{0}^{\pi} \sin x\,dx - \int\limits_{\pi}^{5\pi/4} \sin x\,dx\\[5pt]
&\qquad\quad {}= \Bigl[\ -\cos x\ \Bigr]_0^\pi - \Bigl[\ -\cos x\ \Bigr]_\pi^{5\pi/4}\\[5pt]
&\qquad\quad {}= \Bigl[\ -\cos x\ \Bigr]_0^\pi - \Bigl[\ -\cos x\ \Bigr]_\pi^{5\pi/4}\\[5pt]

Version vom 12:59, 10. Mär. 2009

If we draw the curve y=sinx, we see that the curve lies above the x-axis as far as x= and then lies under the x-axis.

The area of the region between x=0 and x= can therefore be written as

0sinxdx 

whilst the area of the remaining region under the x-axis is equal to

54sinxdx 

(note the minus sign in front of the integral).

The total area becomes

0sinxdx54sinxdx= cosx 0 cosx 54=cos(cos0)cos45(cos)=(1)(1)12(1)=1+112+1=312.


Note: A simple way to obtain the values of cos0, cos and cos(54) is to draw the angles 0, and 54 on a unit circle and to read off the cosine value as the x-coordinate for the corresponding point on the circle.