# Q. Let f: [-1,2] $\rightarrow [0, \infty]$ be a continuous function such that f(x) = f ( 1 - x), $\forall x \in [-1,2].$ If $R_1 = \int \limits_{-1}^2 x f(x) dx$ and $R_2$ are th e area of the region bounded by y = f(x), x = - 1 , x = 2 and the X-axis. Then,

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Solution:

## $R_1 = \int \limits_{-1}^2 x f(x) dx \, \, \, \, \, .......(i)$ Using $\, \, \, \, \int \limits_a^b f(x) dx = \int \limits_a^b f(a + b - x) dx$ $\, \, \, \, \, \, \, R_1 = \int \limits_{-1}^2 (1 - x) f(1 - x) dx$ $\therefore \, \, \, \, \, \, R_1 = \int \limits_{-1}^2 (1 - x) f(x) dx$ $\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, [f(x) = f(1 - x), given]$ Given, $R_2$ is area bounded by f(x), x = - 1 and x = 2. $\therefore \, \, \, \, \, \, \, \, \, \, R_1 = \int \limits_{-1}^2 f(x) dx \, \, \, ...(iii)$ On adding Eqs. (i) and (ii), we get $\, \, \, \, \, \, \, \, \, \, 2R_1 = \int \limits_{-1}^2 f(x) dx\, \, \, \, \, \, \, ....(iv)$ From Eqs. (iii) and (iv), we get $\, \, \, \, \, \, \, \, \, \, 2R_1 = R_2$

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