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Class 12
Math
Calculus
Area
502
150
Find the area of the region enclosed by the curves y= x log x and
y
=
2
x
−
2
x
2
.
502
150
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Similar topics
relations and functions
trigonometric functions
inverse trigonometric functions
application of derivatives
integrals
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Connecting you to a tutor in 60 seconds.
Get answers to your doubts.
Connecting you to a tutor in 60 seconds.
Get answers to your doubts.
View Similar topics ->
relations and functions
trigonometric functions
inverse trigonometric functions
application of derivatives
integrals
View All ↓
Related Questions
Consider the function defined implicity by the equation
y
2
−
2
y
e
s
i
n
−
1
x
+
x
2
−
1
+
[
x
]
+
e
2
s
i
n
−
1
x
=
0
(
where [x] denotes the greatest integer function
)
.
The area of the region bounded by the curve and the line
x
=
−
1
is
The area of the loop of the curve
a
y
2
=
x
2
(
a
−
x
)
is
The area bounded by the two branches of curve
(
y
−
x
)
2
=
x
3
and the straight line x=1 is
The area made by curve
f
(
x
)
=
[
x
]
+
x
−
[
x
]
and x-axis when
0
≤
x
≤
n
(
n
∈
N
)
is equal to { where [x] is greatest integer function}
Match the statements given in List I with the values given in List II.
The area bounded by
y
=
3
−
∣
3
−
x
∣
and
y
=
∣
x
+
1
∣
6
is
Find the area enclosed by
y
=
g
(
x
)
,
x-axis, x=1 and x=37, where g(x) is inverse of
f
(
x
)
=
x
3
+
3
x
+
1
.
If
A
i
is the area bounded by
∣
x
−
a
i
∣
+
∣
y
∣
=
b
i
,
I
∈
N
,
where
a
i
+
1
=
a
i
+
2
3
b
i
and
b
i
+
1
=
2
b
i
,
a
1
=
0
,
=
b
1
=
3
2
,
then
Related Questions
Consider the function defined implicity by the equation
y
2
−
2
y
e
s
i
n
−
1
x
+
x
2
−
1
+
[
x
]
+
e
2
s
i
n
−
1
x
=
0
(
where [x] denotes the greatest integer function
)
.
The area of the region bounded by the curve and the line
x
=
−
1
is
The area of the loop of the curve
a
y
2
=
x
2
(
a
−
x
)
is
The area bounded by the two branches of curve
(
y
−
x
)
2
=
x
3
and the straight line x=1 is
The area made by curve
f
(
x
)
=
[
x
]
+
x
−
[
x
]
and x-axis when
0
≤
x
≤
n
(
n
∈
N
)
is equal to { where [x] is greatest integer function}
Match the statements given in List I with the values given in List II.
The area bounded by
y
=
3
−
∣
3
−
x
∣
and
y
=
∣
x
+
1
∣
6
is
Find the area enclosed by
y
=
g
(
x
)
,
x-axis, x=1 and x=37, where g(x) is inverse of
f
(
x
)
=
x
3
+
3
x
+
1
.
If
A
i
is the area bounded by
∣
x
−
a
i
∣
+
∣
y
∣
=
b
i
,
I
∈
N
,
where
a
i
+
1
=
a
i
+
2
3
b
i
and
b
i
+
1
=
2
b
i
,
a
1
=
0
,
=
b
1
=
3
2
,
then
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