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JEE Main Math Class 12 Application of Integrals Questions
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© examsnet.com
Question : 70
Total: 78
The area of the plane region bounded by the curves
x
+
2
y
2
=
0
and
x
+
3
y
2
=
lis equal to
[2008]
5
3
1
3
2
3
4
3
Validate
Solution:
Given
x
+
2
y
2
=
0
⇒
y
2
=
−
x
2
and
x
+
3
y
2
=
1
⇒
y
2
=
−
1
3
(
x
−
1
)
On solving these two equations we get the points of intersection as (-2,1),(-2,-1)
The required area is ACBDA, given by
A
=
2
{
1
∫
−
2
1
√
3
√
1
−
x
d
x
−
1
√
2
0
∫
−
2
√
−
x
d
x
}
⇒
2
{
1
√
3
[
2
3
(
1
−
x
)
3
∕
2
]
−
2
1
−
1
√
2
[
2
3
(
−
x
)
3
∕
2
]
−
2
0
}
⇒
2
{
[
−
1
√
3
×
2
3
(
0
−
3
3
∕
2
)
]
−
[
−
1
√
2
×
2
3
(
0
−
2
3
∕
2
)
]
}
}
⇒
2
{
2
3
√
3
×
3
√
3
−
1
√
2
×
2
3
.
2
√
2
}
⇒
2
{
2
−
4
3
}
=
2
{
6
−
4
3
}
=
4
3
s
q
.
units
© examsnet.com
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