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Standard XII
Maths
Question
Prove that
∫
a
−
a
f
(
x
)
d
x
=
{
2
∫
a
0
f
(
x
)
d
x
,
i
f
f
i
s
a
n
e
v
e
n
f
u
n
c
t
i
o
n
0
i
f
f
i
s
a
n
o
d
d
f
u
n
c
t
i
o
n
and hence evaluate
∫
1
−
1
sin
7
x
.
cos
4
x
d
x
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Solution
Verified by Toppr
here
f
(
x
)
=
s
i
n
7
x
.
c
o
s
4
x
now
f
(
−
x
)
=
s
i
n
7
(
−
x
)
.
c
o
s
4
(
−
x
)
=
−
s
i
n
7
x
.
c
o
s
4
x
=
−
f
(
x
)
this means
f
(
x
)
is odd function
therefore value of integral is
0
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Similar Questions
Q1
Prove that
∫
a
−
a
f
(
x
)
d
x
=
{
2
∫
a
0
f
(
x
)
d
x
,
i
f
f
i
s
a
n
e
v
e
n
f
u
n
c
t
i
o
n
0
i
f
f
i
s
a
n
o
d
d
f
u
n
c
t
i
o
n
and hence evaluate
∫
1
−
1
sin
7
x
.
cos
4
x
d
x
View Solution
Q2
Prove that
∫
a
0
f
(
x
)
d
x
=
∫
a
0
f
(
a
−
x
)
d
x
and hence evaluate
∫
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/
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0
(
2
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−
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x
)
View Solution
Q3
Match the following
List I
List II
I.
∫
1
−
1
x
|
x
|
d
x
(a)
π
2
II.
∫
π
2
0
(
1
+
log
(
4
+
3
sin
x
4
+
3
cos
x
)
)
d
x
(b)
∫
a
2
0
f
(
x
)
d
x
III.
∫
a
0
f
(
x
)
d
x
(c)
∫
a
0
[
f
(
x
)
+
f
(
−
x
)
]
d
x
IV.
∫
a
−
a
f
(
x
)
d
x
(d)
0
(e)
∫
a
0
f
(
a
−
x
)
d
x
View Solution
Q4
Prove that
∫
a
0
f
(
x
)
d
x
=
∫
a
0
f
(
a
−
x
)
d
x
and hence evaluate
∫
a
0
√
x
√
x
+
√
a
−
x
d
x
.
View Solution
Q5
If
∫
a
0
f
(
x
)
d
x
=
∫
a
0
f
(
a
−
x
)
d
x
, then the value of
∫
π
0
x
f
(
sin
x
)
d
x
=
View Solution