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The parabolas $\mathrm{y}^2=4 \mathrm{x}$ and $\mathrm{x}^2=4 \mathrm{y}$ divide the square region bounded by the lines $\mathrm{x}=$ $4, y=4$ and the coordinate axes. If $S_1, S_2, S_3$ are respectively the areas of these parts numbered from top to bottom; then $\mathrm{S}_1: \mathrm{S}_2: \mathrm{S}_3$ is
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$1: 1: 1$
$1: 1: 1$
$\mathrm{y}^2=4 \mathrm{x}$ and $\mathrm{x}^2=4 \mathrm{y}$ are symmetric about line $\mathrm{y}=\mathrm{x}$
$\Rightarrow$ area bounded between $y^2=4 x$ and $y=x$ is $\int_0^4(2 \sqrt{x}-x) d x=\frac{8}{3}$
$\Rightarrow \mathrm{A}_{\mathrm{s}_2}=\frac{16}{3}$ and $\mathrm{A}_{\mathrm{s}_1}=\mathrm{A}_{\mathrm{s}_3}=\frac{16}{3}$
$\Rightarrow A_{s_1}: A_{s_2}: A_{s_3}:: 1: 1: 1$.
$\Rightarrow$ area bounded between $y^2=4 x$ and $y=x$ is $\int_0^4(2 \sqrt{x}-x) d x=\frac{8}{3}$
$\Rightarrow \mathrm{A}_{\mathrm{s}_2}=\frac{16}{3}$ and $\mathrm{A}_{\mathrm{s}_1}=\mathrm{A}_{\mathrm{s}_3}=\frac{16}{3}$
$\Rightarrow A_{s_1}: A_{s_2}: A_{s_3}:: 1: 1: 1$.
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