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Byju's Answer
Standard XII
Mathematics
Sum of n Terms
Prove that ...
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
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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0
Similar questions
Q.
Prove that
∫
a
0
f
(
x
)
d
x
=
∫
a
0
f
(
a
−
x
)
d
x
and hence evaluate
∫
π
/
2
0
(
2
log
sin
x
−
log
sin
2
x
)
Q.
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
Q.
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
.
Q.
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
=
Q.
a) Prove that
a
∫
a
f
(
x
)
d
x
=
⎧
⎪ ⎪ ⎪
⎨
⎪ ⎪ ⎪
⎩
2
a
∫
0
f
(
x
)
d
x
if
f
(
x
)
is an even function
0
if
f
(
x
)
is an odd function
and hence evaluate
1
∫
−
1
sin
5
x
cos
4
x
d
x
.
b) Prove that
∣
∣ ∣ ∣
∣
a
2
+
1
a
b
a
c
a
b
b
2
+
1
b
c
c
a
c
b
c
2
+
1
∣
∣ ∣ ∣
∣
=
1
+
a
2
+
b
2
+
c
2
.
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