First slide

Evaluation of definite integrals

Question

If $f (a + b + 1 - x) = f (x)$ , for all x, where a and b are fixed positive real numbers, then $\frac{1}{a + b} \int_{a}^{b} x (f (x)) + f (x + 1) d x$ is equal to

Difficult

A

$\int_{a + 1}^{b + 1} f (x) d x$
B

$\int_{a + 1}^{b + 1} f (x + 1) d x$
C

$\int_{a - 1}^{b - 1} f (x + 1) d x$
D

$\int_{a - 1}^{b - 1} f (x) d x$

Solution

$f (x + 1) = f (a + b - x)$
$I = \frac{1}{(a + b)} \int_{a}^{b} x (f (x) + f (x + 1)) d x ........ (1)$
$I = \frac{1}{(a + b)} \int_{a}^{b} (a + b - x) (f (x + 1) + f (x)) d x ...... (2)$
From 1 and 2
$2 I = \int_{a}^{b} (f (x) + f (x + 1)) d x$
$2 I = \int_{a}^{b} f (a + b - x) d x + \int_{a}^{b} f (x + 1) d x$
$2 I = 2 \int_{a}^{b} f (x + 1) d x \Rightarrow I = \int_{a}^{b} f (x + 1) d x$
$= \int_{a + 1}^{b + 1} f (x) d x$

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