Aus Online Mathematik Brückenkurs 2

Version vom 13:36, 21. Okt. 2008 (bearbeiten) Tek (Diskussion | Beiträge) K ← Zum vorherigen Versionsunterschied		Version vom 12:59, 10. Mär. 2009 (bearbeiten) (rückgängig) Tekbot (Diskussion | Beiträge) K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) Zum nächsten Versionsunterschied →
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
-	{{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).
Zeile 15:		Zeile 15:
	The total area becomes		The total area becomes
-	{{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]

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Version vom 12:59, 10. Mär. 2009

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

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

\displaystyle \int\limits_{0}^{\pi} \sin x\,dx

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

\displaystyle -\int\limits_{\pi}^{5\pi/4} \sin x\,dx

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

The total area becomes

\displaystyle \begin{align} & \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\pi - (-\cos 0)\Bigr) - \Bigl( -\cos\frac{5\pi}{4} - (-\cos\pi) \Bigr)\\[5pt] &\qquad\quad {}= \Bigl( -(-1)-(-1) \Bigr) - \Bigl( -\Bigl(-\frac{1}{\sqrt{2}} \Bigr) - \bigl(-(-1)\bigr)\Bigr)\\[5pt] &\qquad\quad {}= 1+1-\frac{1}{\sqrt{2}}+1\\[5pt] &\qquad\quad {}= 3-\frac{1}{\sqrt{2}}\,\textrm{.} \end{align}

Note: A simple way to obtain the values of \displaystyle \cos 0, \displaystyle \cos \pi and \displaystyle \cos (5\pi/4) is to draw the angles \displaystyle 0, \displaystyle \pi and \displaystyle 5\pi/4 on a unit circle and to read off the cosine value as the x-coordinate for the corresponding point on the circle.