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Introduction to definite integration
Question
∫
0
a
a
2
−
x
2
d
x
=
Easy
A
π
a
2
B
π
a
2
2
C
π
a
2
3
D
π
a
2
4
Solution
∫
0
a
a
2
−
x
2
d
x
=
[
x
2
a
2
−
x
2
+
a
2
2
S
i
n
−
1
x
a
]
0
a
=
0
+
a
2
2
×
π
2
−
0
=
π
a
2
4
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