Let f-1-2-0- be a continuous functlon such that fx-f1-x for all x-1-2- - Let R1-12xfx dx- and R2 be the area of the region bounded by y-fx - x-1- x-2- and the x-axis- Then

The correct option is D 2R_1=R_2 R _ 2 =∫ _ 1 ^ 2 f(x)dx R_1=∫_ 1^2xf(x) dx nbsp; nbsp; nbsp; nbsp; =∫ _ 1 ^ 2 (1 x)f(1 x)dx ...

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Byju's Answer

Standard XII

Mathematics

Area between x=g(y) and y Axis

Let f:[-1,2...

Question

Let $f : [- 1, 2] \to [0, \infty)$ be a continuous functlon 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 $y = f (x)$ , $x = - 1, x = 2$ , and the $x$ -axis. Then

R_{1} = 2 R_{2}

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R_{1} = 3 R_{2}

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2 R_{1} = R_{2}

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3 R_{1} = R_{2}

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Solution

The correct option is D $2 R_{1} = R_{2}$
$R_{2} = \int_{- 1}^{2} f (x) d x$
$R_{1} = \int_{- 1}^{2} x f (x) d x$
$= \int_{- 1}^{2} (1 - x) f (1 - x) d x$
$= \int_{- 1}^{2} (1 - x) f (x) d x$
$= \int_{- 1}^{2} f (x) d x - \int_{- 1}^{2} x f (x) d x$
$\Rightarrow R_{1} = R_{2} - R_{1}$
or, $2 R_{1} = R_{2}$

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

The correct option is D 2R1=R2R2=∫2−1f(x)dxR1=∫2−1xf(x)dx =∫2−1(1−x)f(1−x)dx =∫2−1(1−x)f(x)dx =∫2−1f(x)dx−∫2−1xf(x)dx⇒R1=R2−R1or, 2R1=R2

Let $f : [- 1, 2] \to [0, \infty)$ be a continuous functlon 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 $y = f (x)$ , $x = - 1, x = 2$ , and the $x$ -axis. Then