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Some Properties of Definite Integrals
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2 0 1 x dx is equal to
Question:
∫
0
2
∣
1
−
x
∣
d
x
\int\limits^2_0 |1 -x| dx
0
∫
2
∣1
−
x
∣
d
x
is equal to
BITSAT - 2011
BITSAT
Updated On:
Jan 30, 2025
0
1
3
2
\frac{3}{2}
2
3
1
2
\frac{1}{2}
2
1
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B
Solution and Explanation
∫
0
2
∣
1
−
x
∣
d
x
=
∫
0
1
(
1
−
x
)
d
x
+
∫
1
2
(
x
−
1
)
d
x
\int\limits_{0}^{2}|1-x| d x=\int\limits_{0}^{1}(1-x) d x+\int\limits_{1}^{2}(x-1) d x
0
∫
2
∣1
−
x
∣
d
x
=
0
∫
1
(
1
−
x
)
d
x
+
1
∫
2
(
x
−
1
)
d
x
=
[
x
−
x
2
2
]
0
1
+
[
x
2
2
−
x
]
1
2
=\left[x-\frac{x^{2}}{2}\right]_{0}^{1}+\left[\frac{x^{2}}{2}-x\right]_{1}^{2}
=
[
x
−
2
x
2
]
0
1
+
[
2
x
2
−
x
]
1
2
=
1
−
1
2
+
[
2
−
2
−
(
1
2
−
1
)
]
=1-\frac{1}{2}+\left[2-2-\left(\frac{1}{2}-1\right)\right]
=
1
−
2
1
+
[
2
−
2
−
(
2
1
−
1
)
]
=
1
2
+
1
2
=
1
=\frac{1}{2}+\frac{1}{2}=1
=
2
1
+
2
1
=
1
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Let
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a
⃗
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b
⃗
,
c
⃗
\vec{a}, \vec{b}, \vec{c}
a
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b
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c
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A
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B
,
C
A, B, C
A
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B
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D
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BC
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B
E
BE
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A
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AC
A
C
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F
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BF
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:
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F
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α
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3
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4
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(3,4,5)
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3
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(4,6,3)
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(-1,2,4)
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1
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2
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4
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1
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0
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(1,0,5)
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1
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0
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6
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m
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2
n
+
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l
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