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Math 213 — Quiz 7
Name
Solution
1. Show that 0
≤
RR
R
sin(
x
+
y
)
dA
≤
1 where
R
= [0
,
1]
×
[0
,
1].
When (
x,y
)
∈
R
, i.e. 0
≤
x
≤
1 and 0
≤
y
≤
1, we have 0
≤
x
+
y
≤
2. Since
0
≤
sin(
z
)
≤
1 when 0
≤
z
≤
π
and
π
≥
2, it follows that 0
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Unformatted text preview: ≤ sin( x + y ) ≤ 1 for ( x,y ) ∈ R . Therefore, 0 = ZZ R dA ≤ ZZ R sin( x + y ) dA ≤ ZZ R 1 dA = 1 as claimed....
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This note was uploaded on 05/09/2008 for the course MATH 2130 taught by Professor Dorais during the Spring '08 term at Cornell University (Engineering School).
 Spring '08
 DORAIS
 Math, Calculus

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