de-CH
utf-8
math math-format
Gebietsintegral
int2-01-01
multiple
19440
randRangeExclude(-8,8,[0,1]) randRangeExclude(-8,8,[0,1,A]) randRange(1,8) randRange(1,8)

Gegeben sei die Funktion f: \mathbb R^2 \to \mathbb R mit f(x,y) = A x + B y.

Berechnen Sie das Integral \displaystyle \int \int_{{\color{orange}D}} f(x,y) dA über dem Gebiet \color{orange}D.



graphInit({ range: [[-9, 9],[-9, 9]], scale: [20,20], tickStep: [1,1], gridStep: [1,1], labelStep: [2,2], gridOpacity: 0.1, axisOpacity: 0.8, tickOpacity: 0.6, labelOpacity: 0.8 }); label( [ 0, 9.5 ], "y", "above" ); label( [9.5,0 ], "x", "right" ); line( [0, 0], [X/2, 0], { stroke: ORANGE } ); line( [X/2, 0], [X, 0], { stroke: ORANGE } ); line( [X,0], [X, Y], { stroke: ORANGE } ); line( [X, Y], [0, Y], { stroke: ORANGE } ); line( [0,Y], [0, 0], { stroke: ORANGE } );
a \displaystyle \int \int_{{\color{orange}D}} f(x,y) dA = 0.5*A*Y*X*X+B*Y*Y*X*0.5

Es ist \displaystyle \int\int_{\orange{D}} f(x,y) dA = \int_0^{X} \int_0^{Y} (negParens(A)\cdot x + negParens(B)\cdot y) dy dx.

Die innere Integration ist

\displaystyle \int_0^{Y} (negParens(A)\cdot x + negParens(B)\cdot y) dy = (negParens(A)\cdot x) \cdot y + fractionReduce(B,2)\cdot y^2\bigg|_0^{Y} = negParens(A*Y)\cdot x + fractionReduce(B*Y*Y,2).

Damit erhalten wir für die äussere Integration:

\displaystyle \int_0^{X}\left(negParens(A*Y)\cdot x + fractionReduce(B*Y*Y,2) \right) dx = negParens(fractionReduce(A*Y,2))\cdot x^2 + fractionReduce(B*Y*Y,2)\cdot x \bigg|_0^{X} = fractionReduce(A*Y*X*X+B*Y*Y*X,2).