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Byju's Answer
Standard XII
Physics
Definite Integrals
Let ddxFx= es...
Question
Let
d
d
x
F
(
x
)
=
(
e
sin
x
x
)
, where
x
>
0
. If
4
∫
1
3
x
e
sin
x
3
d
x
=
F
(
k
)
−
F
(
1
)
, then the possible value of
k
is
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Solution
For
4
∫
1
3
x
e
sin
x
3
d
x
Putting
x
3
=
t
⇒
3
x
2
d
x
=
d
t
When
x
=
1
⇒
t
=
1
;
x
=
4
⇒
t
=
64
Now,
4
∫
1
3
x
e
sin
x
3
d
x
=
64
∫
1
e
sin
t
t
d
t
=
64
∫
1
d
d
t
F
(
t
)
⋅
d
t
=
F
(
64
)
−
F
(
1
)
∴
k
=
64
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0
Similar questions
Q.
Let
d
d
x
F
(
x
)
=
(
e
s
i
n
x
x
)
,
x
>
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f
∫
4
1
3
x
e
s
i
n
x
3
d
x
=
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−
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Q.
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d
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x
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x
>
0
. If
∫
4
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2
e
sin
x
2
x
d
x
=
F
(
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−
F
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)
, then the possible value of
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Q.
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)
=
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s
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3
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3
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k
)
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then one possible value of k is
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d
x
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x
)
=
e
s
m
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x
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x
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.
lf
∫
4
1
3
x
e
s
m
x
3
d
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=
F
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