First slide

Evaluation of definite integrals

Question

$Let f be a real-valued function satisfying f (x) + f (x + 4) = f (x + 2) + f (x + 6) then \int_{x}^{x + 8} f (t) d t is$

Moderate

A

A function in terms of x
B

constant function
C

Periodic function
D

Depends on x

Solution

$Given that f (x) + f (x + 4) = f (x + 2) + f (x + 6) - - - - (1)$
$Replacing x by x + 2, we get$
$f (x + 2) + f (x + 6) = f (x + 4) + f (x + 8) - - - - (2)$
$From equations (1) and (2), we get f (x) = f (x + 8) - - - - (3)$
$or \int_{x}^{x + 8} f (t) d t = \int_{0}^{8} f (t) d t = constant$

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