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NDA Math Definite Integrals
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© examsnet.com
Question : 12
Total: 62
What is
a
∫
0
f
(
a
−
x
)
f
(
x
)
+
f
(
a
−
x
)
d
x
equal to?
[NDA-I 2021]
a
2
a
0
a
2
Validate
Solution:
👈: Video Solution
Concept:
Definite Integrals:
b
∫
a
f
(
x
)
d
x
=
b
∫
a
f
(
a
+
b
−
x
)
d
x
If
f
(
x
)
=
f
(
2
a
−
x
)
, then
2
a
∫
0
f
(
x
)
d
x
=
2
a
∫
0
f
(
x
)
d
x
A function
f
(
x
)
is:
- Even
, if
f
(
−
x
)
=
f
(
x
)
. And
a
∫
−
a
f
(
x
)
d
x
=
2
a
∫
0
f
(
x
)
d
x
- Odd
,if
f
(
−
x
)
=
−
f
(
x
)
. And
∫
−
a
a
f
(
x
)
d
x
=
0
.
- Periodic
, if
f
(
n
p
±
x
)
=
f
(
x
)
, for some number
p
and
n
∈
Z
.
Calculation:
We know that
b
∫
a
f
(
x
)
d
x
=
b
∫
a
f
(
a
+
b
−
x
)
d
x
∴
I
=
∫
0
a
f
(
a
−
x
)
f
(
x
)
+
f
(
a
−
x
)
d
x
=
∫
0
a
f
[
(
a
+
0
)
−
(
a
−
x
)
]
f
[
(
a
+
0
)
−
x
]
+
f
[
(
a
+
0
)
−
(
a
−
x
)
]
d
x
=
∫
0
a
f
(
x
)
f
(
a
−
x
)
+
f
(
x
)
d
x
And,
2
I
=
∫
0
a
f
(
a
−
x
)
f
(
x
)
+
f
(
a
−
x
)
d
x
+
∫
0
a
f
(
x
)
f
(
a
−
x
)
+
f
(
x
)
d
x
=
a
∫
0
1
d
x
=
a
.
⇒
I
=
a
2
© examsnet.com
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