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Unformatted text preview: x − 4) − 3y } dy dx. Now Explanation:
After integration with respect to y we see
that
1 5
3xy − y 2
2 I=
0 1 =
0 = x2
0 {4(x − 4) − 3y } dy 0 dx = 5
3 x3 − x4 d x
2
34 15
x− x
4
2 x −4 0 . x −4
0 . Consequently
6 1 3
4(x − 4)y − y 2
2 I=
4 5
(x − 4)2 dx =
2 5
(x − 4)3
6 6
4 , rogers (grr459) – HW13 – rusin – (55220) 6
y and so
I= 011 20
3 4 . 10.0 points Reverse the order of integration in the integral
x
2x 2 2 f (x, y ) dy dx , I=
0 0 but make no attempt to evaluate either integral.
2 4 1. I = f (x, y ) dx dy
y 0 Integration is taken ﬁrst with respect to y for
ﬁxed x along the dashed vertical line.
To change the order of integration, now ﬁx
y and let x vary along the solid horizontal line
in
y y 4 2. I = f (x, y ) dx dy
0 4 2
4 2 3. I = f (x, y ) dx dy correct
y/2 0 2y 2 4. I = f (x, y ) dx dy
0 0
y /2 4 5. I = x f (x, y ) dx dy
0
2 4 6. I = f (x, y ) dx dy
0 2 0 2y Explanation:
The region of integration is the set of all
points
(x, y ) : 0 ≤ y ≤ 2x , 0 ≤ x ≤ 2
in the plane bounded by the xaxis and the
graphs of
y = 2x , x = 2. This is the shaded region in Since the equation of the slant line is x = y/2,
integration in x is along the line from (y/2, y )
to (2, y ) for ﬁxed y , and then from y = 0 to
y = 4.
Consequently, after changing the order of
integration,
4 2 I= f (x, y ) dx dy
0 . y/2 keywords: double integral, reverse order integration, linear function,...
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This note was uploaded on 09/22/2013 for the course M 408 D taught by Professor Textbookanswers during the Fall '07 term at University of Texas.
 Fall '07
 TextbookAnswers

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