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Standard XII
Mathematics
First Fundamental Theorem of Calculus
The value of ...
Question
The value of integral
∫
e
6
1
[
l
o
g
x
3
]
d
x
, where [.] denotes the greatest integer function, is
A
0
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B
e
6
−
e
3
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C
e
6
+
e
3
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D
e
3
−
e
6
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Solution
The correct option is
B
e
6
−
e
3
W
h
e
n
1
<
x
<
e
3
,
[
l
o
g
x
3
]
=
0
A
n
d
w
h
e
n
e
3
<
x
<
e
6
,
[
l
o
g
x
3
]
=
1
∴
∫
e
6
1
[
l
o
g
x
3
]
d
x
=
∫
e
3
1
[
l
o
g
x
3
]
d
x
+
∫
e
6
e
3
[
l
o
g
x
3
]
d
x
=
∫
e
3
1
0
d
x
+
∫
e
6
e
3
1
d
x
=
(
e
6
−
e
3
)
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Q.
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∫
e
6
1
[
l
o
g
x
3
]
d
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, where [.] denotes the greatest integer function, is
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