Let S be the region bounded by the curves y=x | vedclass

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(B) The region $S$ is bounded by $y=x^{3}$ and $y^{2}=x$ in the first quadrant. The intersection points are $(0,0)$ and $(1,1)$.The total area $S$ is

Vedclass · Accepted Answer

$19$

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(B) The region $S$ is bounded by $y=x^{3}$ and $y^{2}=x$ in the first quadrant. The intersection points are $(0,0)$ and $(1,1)$.The total area $S$ is given by:$S = \int_{0}^{1} (\sqrt{x} - x^{3}) dx$$= \left[ \frac{2}{3}x^{3/2} - \frac{x^{4}}{4} ight]_{0}^{1} = \frac{2}{3} - \frac{1}{4} = \frac{5}{12}$.The curve $y=2x$ (for $x>0$) intersects $y^{2}=x$ at $4x^{2}=x$,so $x=1/4$ (since $x 
eq 0$).The area $R_{1}$ is bounded by $y=\sqrt{x}$ and $y=2x$ from $x=0$ to $x=1/4$:$R_{1} = \int_{0}^{1/4} (\sqrt{x} - 2x) dx$$= \left[ \frac{2}{3}x^{3/2} - x^{2} ight]_{0}^{1/4} = \frac{2}{3}(\frac{1}{8}) - \frac{1}{16} = \frac{1}{12} - \frac{1}{16} = \frac{4-3}{48} = \frac{1}{48}$.Since $S = R_{1} + R_{2}$,we have $R_{2} = S - R_{1} = \frac{5}{12} - \frac{1}{48} = \frac{20-1}{48} = \frac{19}{48}$.Therefore,$\frac{R_{2}}{R_{1}} = \frac{19/48}{1/48} = 19$.

Let $S$ be the region bounded by the curves $y=x^{3}$ and $y^{2}=x$. The curve $y=2|x|$ divides $S$ into two regions of areas $R_{1}$ and $R_{2}$. If $\max \{R_{1}, R_{2}\}=R_{2}$,then $\frac{R_{2}}{R_{1}}$ is equal to

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