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Curves
=
r 2sinθ
and
=
r 2sin2θ
,
≤
≤
0
θ π2
Area of the region outside the first curve and inside the second curve:
The point of intersection of both curves:
=
=>
=
2sinθ 2sin2θ
θ π3
For given case,
≤
≤
0
θ π3
Required area is:
=
A 0π32sinθ2sin2θrdrdθ
=

420π3sin22θ sin2θdθ
=

 
20π31 cos4θ2 1 cos2θ2dθ
=

0π3cos2θ cos4θdθ
=

sin2θ2 sin4θ40π3
=
A
338
2.
Convergence of series:
= ∞
n 1 15n
=
+
+
+…+
Sn 15 110 115
15n
And,
+ 
=
( + )+
( + )+…+
( + )>
( + )+
( + )+…+
( + )
Sn p Sn 15 n 1 15 n 2
15 n p 15 n p 15 n p
15 n p
Now,
+ =
=
p5n p 110 when p n
i.e.

>
S2n Sn 110
Thus for
<
ε 110
, we cannot have

<
S2n Sn ε
for all
≥
n N
.
Hence series is
divergent
.
3.
Series:
= ∞ +
+
n 1 n 12nn 2
=
+
+
an
n 12nn 2
Now,
→∞ +
+ =
limn
n 12nn 2 1
which is not equal to 0.
Hence, series is
divergent.
Sum = DNE
4.
Series:
= ∞
+
n 1 4n 15n
By ratio test,
→∞
+
<
limn
an 1an 1
for convergence.
Now,
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 Spring '11
 Prof.Ramachandran

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