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Lösung 2.1:1d

Aus Online Mathematik Brückenkurs 2

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The modulus function, x, strips x of its sign, e.g.

−5=5, 3=3 and −=.

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

If we draw a graph of y=x it will consist of two parts. For x0 we have y=x, and for x0 we have \displaystyle y=-x\,.

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

This region consists of two triangles and we therefore obtain

\displaystyle \int\limits_{-1}^{2} |x|\,dx = \frac{1}{2}\cdot 1\cdot 1 + \frac{1}{2}\cdot 2\cdot 2 = \frac{5}{2}\,\textrm{.}
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.1:1d“