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Q.

Suppose for every integer n,  nn+1f(x)dx=n2 then the value of -22f(x)dx=

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a

12

b

20

c

6

d

19

answer is A.

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Detailed Solution

nn+1f(x) dx=n2 -22f(x)dx=-2-1f(x)+-10f(x)+01f(x)+12f(x) dx =(-2)2+(-1)2+02+12 =4+1+1=6

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Suppose for every integer n,  ∫nn+1f(x)dx=n2 then the value of ∫-22f(x)dx=