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

∫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

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

∫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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$S u p p o s e f o r e v e r y i n t e g e r n, \int_{n}^{n + 1} f (x) d x = n^{2} t h e n t h e v a l u e o f \int_{- 2}^{2} f (x) d x =$

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Detailed Solution

$\int_{n}^{n + 1} f (x) dx = n^{2} \int_{- 2}^{2} f (x) dx = \int_{- 2}^{- 1} f (x) + \int_{- 1}^{0} f (x) + \int_{0}^{1} f (x) + \int_{1}^{2} f (x) dx = {(- 2)}^{2} + {(- 1)}^{2} + 0^{2} + 1^{2} = 4 + 1 + 1 = 6$

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