Lösung 2.1:4a

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If we draw the curve \displaystyle y=\sin x, we see that the curve lies above the \displaystyle x -axis as far as \displaystyle x=\pi and then lies under the \displaystyle 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 \displaystyle 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 \\ & =\left[ -\cos x \right]_{0}^{\pi }-\left[ -\cos x \right]_{\pi }^{{5\pi }/{4}\;} \\ & =\left( -\cos \pi -\left( -\cos 0 \right) \right)-\left( -\cos \frac{5\pi }{4}-\left( -\cos \pi \right) \right) \\ & =\left( -\left( -1 \right)-\left( -1 \right) \right)-\left( -\left( -\frac{1}{\sqrt{2}} \right)-\left( -\left( -1 \right) \right) \right) \\ & =1+1-\frac{1}{\sqrt{2}}+1=3-\frac{1}{\sqrt{2}} \\ \end{align}


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