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Byju's Answer
Standard XII
Mathematics
Area between Two Curves
Sketch the re...
Question
Sketch the region bounded by the curves
y
=
√
5
−
x
2
and
y
=
|
x
−
1
|
and find its area using integration.
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Solution
The given curves are :
x
2
+
y
2
=
5
[
∵
y
=
√
5
−
x
2
⇒
y
2
=
5
−
x
2
⇒
x
2
+
y
2
=
5
]
and
y
=
{
1
−
x
,
i
f
x
<
1
x
−
1
,
i
f
x
>
1
The required region is shown as shaded in the following figure :
y
=
x
−
1
meets
x
2
+
y
2
=
5
at
B
(
2
,
1
)
.
y
=
1
−
x
meets
x
2
+
y
2
=
5
at
C
(
−
1
,
2
)
y
=
x
−
1
and
y
=
1
−
x
meet at
A
(
1
,
0
)
.
∴
Reqd. area = area (MCBLM)- area (CMAC)- area (ALBA)
=
∫
2
−
1
√
5
−
x
2
d
x
−
∫
1
−
1
(
1
−
x
)
d
x
−
∫
2
1
(
x
−
1
)
d
x
=
[
x
√
5
−
x
2
2
+
5
2
sin
−
1
x
√
5
]
2
−
1
−
[
x
−
x
2
2
]
1
−
1
−
[
x
2
2
−
x
]
2
1
=
[
(
1
+
5
2
sin
−
1
2
√
5
)
−
(
−
1
2
×
2
+
5
2
sin
−
1
(
−
1
√
5
)
)
]
−
[
(
1
−
1
2
)
−
(
−
1
−
1
2
)
]
−
[
(
2
−
2
)
−
(
1
2
−
1
)
]
=
1
+
5
2
sin
−
1
2
√
5
+
1
−
5
2
sin
−
1
(
−
1
√
5
)
−
2
−
1
2
=
−
1
2
+
5
2
[
sin
−
1
2
√
5
−
sin
−
1
(
−
1
√
5
)
]
=
−
1
2
+
5
2
[
sin
−
1
2
√
5
+
sin
−
1
(
1
√
5
)
]
∵
sin
−
1
x
+
sin
−
1
y
=
sin
−
1
[
x
.
√
1
−
y
2
]
+
sin
−
1
[
y
.
√
1
−
x
2
]
x
≥
0
,
y
≥
0
x
2
+
y
2
≤
0
=
−
1
2
+
5
2
[
sin
−
1
(
2
√
5
√
1
−
1
5
+
1
√
5
√
1
−
4
5
)
]
=
−
1
2
+
5
2
[
sin
−
1
(
4
5
+
1
5
)
]
=
−
1
2
+
5
2
[
sin
−
1
1
]
=
−
1
2
+
5
2
×
π
2
=
(
−
1
2
+
5
π
4
)
sq. units
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