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NDA Math Definite Integrals
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© examsnet.com
Question : 45
Total: 62
1
∫
−
1
x
|
x
|
is equal to
[NDA II 2015]
0
2
3
2
−
2
Validate
Solution:
1
∫
−
1
x
|
x
|
d
x
=
0
∫
−
1
x
|
x
|
d
x
+
1
∫
0
x
|
x
|
d
x
=
0
∫
−
1
x
(
−
x
)
d
x
+
1
∫
0
x
⋅
x
d
x
=
−
0
∫
−
1
x
2
d
x
+
1
∫
0
x
2
d
x
=
−
[
x
3
3
]
−
1
0
+
[
x
3
3
]
0
1
=
−
[
0
−
(
−
1
)
3
3
]
+
1
3
[
(
1
)
3
−
(
0
)
3
]
=
−
1
3
+
1
3
=
0
© examsnet.com
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