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Methods of integration

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Question

$Let f : R \to R be a continuous function such that f (x) + f (x + 1) = 2, for all x \in R . If$ $I_{1} = \int_{0}^{8} f (x) d x and I_{2} = \int_{- 1}^{3} f (x) d x, then the value of I_{1} + 2 I_{2} is equal to$

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Solution

$f (x) + f (x + 1) = 2 a n d f (x + 1) + f (x + 2) = 2 \Rightarrow f (x + 2) = f (x)$
$T h e f u n c t i o n f (x) i s p e r i o d i c w i t h p e r i o d i c i t y 2$
$I_{1} = 4 \int_{0}^{2} f (x) d x . s u b s t i t u t e x = t + 1 i n I_{2}$
$I_{2} = \int_{- 2}^{2} f (t + 1) d t = 2 \int_{0}^{2} f (x + 1) d t I_{1} + 2 I_{2} = 4 \int_{0}^{2} f (x) + f (x + 1) d x = 4 \times 4 = 16$

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