Lösung 2.1:4a
Aus Online Mathematik Brückenkurs 2
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.