Let f:[−1,2]→[0,∞] be a continuous function such that f(x)=f(1−x) for all x∈[−1,2]. Let R1=∫−12 xf(x)dx., and R2 be the area of the region bounded by y=f(x),x=−1 and x=2 and the x-axis. Then

R1=∫−12 xf(x)dx=∫−12 (2+(−1)−x)f(2+(−1)−x)dx=∫−12 (1−x)f(1−x)dx=∫−12 (1−x)f(x)dx⇒2R1=∫−12 f(x)dx=R2.

Let f:[−1,2]→[0,∞] be a continuous function such that f(x)=f(1−x) for all x∈[−1,2]. Let R1=∫−12 xf(x)dx., and R2 be the area of the region bounded by y=f(x),x=−1 and x=2 and the x-axis. Then

R1=∫−12 xf(x)dx=∫−12 (2+(−1)−x)f(2+(−1)−x)dx=∫−12 (1−x)f(1−x)dx=∫−12 (1−x)f(x)dx⇒2R1=∫−12 f(x)dx=R2.

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Let $f : [- 1, 2] \to [0, \infty]$ be a continuous function such that $f (x) = f (1 - x)$ for all $x \in [- 1, 2]$ . Let $R_{1} = \int_{- 1}^{2} x f (x) d x$ ., and $R_{2}$ be the area of the region bounded by $y = f (x), x = - 1$ and $x = 2$ and the x-axis. Then

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$R_{1} = 2 R_{2}$

$R_{1} = 3 R_{2}$

$2 R_{1} = R_{2}$

$3 R_{1} = R_{2}$

answer is C.

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Detailed Solution

$\begin{array}{l} R_{1} = \int_{- 1}^{2} x f (x) d x = \int_{- 1}^{2} (2 + (- 1) - x) f (2 + (- 1) - x) d x \\ = \int_{- 1}^{2} (1 - x) f (1 - x) d x = \int_{- 1}^{2} (1 - x) f (x) d x \\ \Rightarrow 2 R_{1} = \int_{- 1}^{2} f (x) d x = R_{2} . \end{array}$