Page 42 - MODUL Kalkulus Lanjut

sehingga diperoleh bentuk:

f ( x)
2
b
 f ( x, y) dA =   f ( x, y) dydx
R a f ( x)
1

Contoh:

1 x
  xy 2 dydx
1.
0 x 2

Penyelesaian:

1 x 1  xy 3  x
  xy 2 dydx =   dx


0 x 2 0  3  x 2
1  (x 2 ) 3 x (x ) 3 
x

= .   − dx

0  3 3 
1 1
= .  (x 7 − x 4 ) dx
3 0

1  x x  1
=  − 
3  8 5 0 

1  1 1 
=  − 
3  8 5 

1
= −
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2 3y 2  x 2  3y

2.   (x + )dxdy =   − xy  dy

y

1 y 1  2  y

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