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A4-format til udskrift. - Aarhus Universitet

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Eksempel 5<br />

<br />

R<br />

sin(x)cos(y)dA =<br />

2. ITERERET INTEGRAL 89<br />

=<br />

=<br />

π<br />

2<br />

0<br />

π<br />

2<br />

0<br />

π<br />

2<br />

π<br />

2<br />

0<br />

cos(y)<br />

sin(x)cos(y)dxdy<br />

π<br />

2<br />

0<br />

sin(x)dx<br />

sin(x)dx dy<br />

π<br />

2<br />

0<br />

0<br />

= [−cos(x)] π<br />

π<br />

2 2<br />

0 [sin(y)] 0<br />

= (−0 − (−1))(1 − 0)<br />

= 1<br />

cos(y)dy<br />

2.28. Opgave ☞ [S] 12.2 Iterated integrals<br />

Øvelse 16<br />

x<br />

R = [0,1] × [0,1], f(x,y) = xe xy<br />

z<br />

2.29. Opgave løst<br />

Øvelse 16<br />

☞ [S] 12.2 Iterated integrals<br />

<br />

xe xy 1 1<br />

dA = xe xy dydx<br />

R<br />

0 0<br />

1<br />

= [e<br />

0<br />

xy ] y=1<br />

y=0 dx<br />

1<br />

= (e<br />

0<br />

x − e 0 )dx<br />

= [e x − x] 1<br />

0<br />

= e − 2<br />

2.30. Test Fubini ☞ [S] 12.2 Iterated integrals<br />

Test<br />

Lad f(x,y) = xlny, R = [0,2] ×[1,2]. Der gælder <br />

R xlny dA = 2<br />

0 x dx 2<br />

lny dy. 1<br />

Afkryds:<br />

ja<br />

<br />

nej<br />

y

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