de-CH
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math math-format
Application of Integral Calculation Rules
i-06-02
number
200
randRange(1,10) randRangeNonZero(-10,10) function( t ) { return [t, sin(t)]; } function( t ) { return [t, 0]; } function( t ) { return [t, -sin(t)]; }

Let f an odd function with

\displaystyle\int_0^{U} f(x)\; dx = I.

Calculate

\displaystyle\int_{-U}^0 f(x) \; dx.

-I

The graph of an odd function (z.B. \pm \sin) ist symmetric about the origin.

graphInit({ range: [ 1.2 * PI, 1.2 ], scale: [ 40, 40 ], gridStep: [ PI/2, .5 ], tickStep: [ 2, 2 ], labelStep: [ 2, 1 ], unityLabels: true, }); // draw curve style({ stroke: BLUE, strokeWidth: 2.2 }, function() { plot( function( x ) { return sin (x); }, [ -PI+0.5, PI-0.5 ] ); });
graphInit({ range: [ 1.2 * PI, 1.2 ], scale: [ 40, 40 ], gridStep: [ PI/2, .5 ], tickStep: [ 2, 2 ], labelStep: [ 2, 1 ], unityLabels: true, }); // draw curve style({ stroke: BLUE, strokeWidth: 2.2 }, function() { plot( function( x ) { return -sin (x); }, [ -PI+0.5, PI-0.5 ] ); });

and the areas above- und below the x-axis are each of equal size.

style({ fill: BLUE, stroke: BLUE, opacity: 0.5 }, function() { plotParametric( fn1, [-PI+0.5, 0], 1, fn2); }); style({ fill: ORANGE, stroke: BLUE, opacity: 0.5 }, function() { plotParametric( fn1, [ 0, PI-0.5], 1, fn2); });
style({ fill: ORANGE, stroke: BLUE, opacity: 0.5 }, function() { plotParametric( fn2, [-PI+0.5, 0], 1, fn3); }); style({ fill: BLUE, stroke: BLUE, opacity: 0.5 }, function() { plotParametric( fn3, [ 0, PI-0.5], 1, fn2); });

From the point-symmetric graph, we see:

\displaystyle \pm \left( \int_{\color{red}{-U}}^{\color{blue}{0}} f(x)\; dx \right) = Left area = Right area = \mp \left( \displaystyle \int_{0}^{U} f(x)\; dx \right) .

Thus:

\displaystyle \int_{\color{red}{-U}}^{\color{blue}{0}} f(x)\; dx = -I.