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Byju's Answer
Standard XII
Mathematics
First Fundamental Theorem of Calculus
∫02[x2]dx is ...
Question
∫
2
0
[
x
2
]
d
x
is (where [.] is greastest integral function
A
2
−
√
2
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B
2
+
√
2
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C
√
2
−
1
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D
−
√
2
−
√
3
+
5
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Solution
The correct option is
D
−
√
2
−
√
3
+
5
∫
2
0
[
x
2
]
d
x
=
∫
1
0
[
x
2
]
d
x
+
∫
√
2
0
[
x
2
]
d
x
+
∫
√
3
√
2
[
x
2
]
d
x
+
∫
2
√
3
[
x
2
]
d
x
=
∫
1
0
0
d
x
+
∫
√
2
0
1
d
x
+
∫
√
3
√
2
2
d
x
+
∫
2
√
3
3
d
x
=
√
2
−
1
+
2
√
3
−
2
√
2
+
6
−
3
√
3
=
5
−
√
3
−
√
2
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Similar questions
Q.
∫
2
0
[
x
2
]
d
x
is (where [.] is greastest integral function
Q.
The value of the integral
∫
2
0
∣
∣
x
2
−
1
∣
∣
d
x
is
Q.
∫
2
0
(
[
x
]
2
−
[
x
2
]
)
d
x
is equal to (where
[
.
]
denotes the greatest integer function)
Q.
The value of integral
∫
2
0
(
∣
∣
x
2
−
3
x
+
2
∣
∣
+
|
cos
x
|
)
d
x
, is equal to
Q.
The value of
∫
2
0
[
x
2
−
1
]
d
x
dx, where [x] denotes the greatest integer function is given by:
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