Lösung 2.1:1d

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The modulus function, \displaystyle \left| x \right|, strips \displaystyle x of its sign, e.g.


\displaystyle \left| -5 \right|=5, \displaystyle \left| 3 \right|=3 and \displaystyle \left| -\pi \right|=\pi


For positive values of \displaystyle x, the modulus function has no effect, since \displaystyle \left| x \right|=x, but for negative \displaystyle x the modulus function changes the sign of \displaystyle x, i.e. \displaystyle \left| -x \right|=x (remember that \displaystyle x is negative and therefore \displaystyle -x is positive).

If we draw a graph of \displaystyle y=\left| x \right| it will consist of two parts. For \displaystyle x\ge 0 we have \displaystyle y=x, and for \displaystyle x\le 0 we have \displaystyle y=-x




The value of the integral is the area of the region under the graph \displaystyle y=\left| x \right| and between \displaystyle x=-1 and \displaystyle x=2.



This region consists of two triangles and we therefore obtain

\displaystyle \int\limits_{-1}^{2}{\left| x \right|}\,dx=\frac{1}{2}\centerdot 1\centerdot 1+\frac{1}{2}\centerdot 2\centerdot 2=\frac{5}{2}