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Question

The value of integral 44x20211+x2022dx, is

A
0
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B
2
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C
4
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D
8
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Solution

The correct option is A 0
Let I=44x20211+x2022dx
We know that,
aaf(x) dx=⎪ ⎪ ⎪⎪ ⎪ ⎪2a0f(x) dx, if f(x) is even0, if f(x) is odd
I=0
(x20211+x2022 is an odd function.)

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