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Step-by-Step Solution
Step 1: Understand the Curves and the Region S
We are given two curves:
Curve 1: \(y = x^3\)
Curve 2: \(y^2 = x\), or equivalently \(y = \pm\sqrt{x}\).
The region \(S\) is the area enclosed between these curves. Since \(y^2 = x\) implies \(x \ge 0\), the relevant portion of \(y = x^3\) is also restricted to \(x \ge 0\). Over \(x \ge 0\), \(x^3 \ge 0\). Hence the portion of \(y^2 = x\) that bounds the region in the first quadrant is \(y = \sqrt{x}\).
The points of intersection of \(y = x^3\) and \(y = \sqrt{x}\) occur where:
\[
x^3 = \sqrt{x}
\quad\Longrightarrow\quad
x^6 = x
\quad\Longrightarrow\quad
x^6 - x = 0
\quad\Longrightarrow\quad
x(x^5 - 1) = 0
\quad\Longrightarrow\quad
x = 0 \text{ or } x^5 = 1 \ (\text{giving } x = 1).
\]
Hence the curves intersect at \((0,0)\) and \((1,1)\). Therefore, the region \(S\) spans \(x\) from \(0\) to \(1\), and for each \(x\) in this interval, \(y\) goes from \(x^3\) (lower boundary) up to \(\sqrt{x}\) (upper boundary).
Step 2: Identify the Dividing Curve \(y = 2|x|\)
Since the region of interest is \(x \ge 0\), \(|x| = x\). Hence on \([0,1]\), the line is:
\[
y = 2x.
\]
We want to see how \(y = 2x\) divides the region \(S\). We check intersections with the bounding curves:
Intersection with \(y = x^3\):
\[
x^3 = 2x
\quad\Longrightarrow\quad
x^3 - 2x = 0
\quad\Longrightarrow\quad
x(x^2 - 2) = 0
\quad\Longrightarrow\quad
x=0 \ (\text{or } x=\pm\sqrt{2}).
\]
Only \(x=0\) lies in \([0,1]\).
Intersection with \(y = \sqrt{x}\):
\[
\sqrt{x} = 2x
\quad\Longrightarrow\quad
4x^2 = x
\quad\Longrightarrow\quad
x(4x - 1) = 0
\quad\Longrightarrow\quad
x=0 \text{ or } x=\tfrac{1}{4}.
\]
Within \([0,1]\), the relevant intersection is \(x=\tfrac{1}{4}\).
So the line \(y = 2x\) meets the region’s top boundary at \(x=0\) and \(x=\tfrac{1}{4}\). For \(x \in (0,\tfrac{1}{4})\), we have
\[
x^3 < 2x < \sqrt{x},
\]
while for \(x \in [\tfrac{1}{4},1]\), \(2x\) becomes larger than \(\sqrt{x}\).
Step 3: Total Area of the Region \(S\)
First, let us find the total area of \(S\), given by
\[
\int_{0}^{1}
\bigl[\sqrt{x} \;-\; x^3\bigr]\,dx.
\]
Compute separately:
\(\displaystyle \int_{0}^{1} \sqrt{x}\,dx
= \int_{0}^{1} x^{\tfrac{1}{2}}\;dx
= \left[\frac{2}{3}x^{\tfrac{3}{2}}\right]_{0}^{1}
= \frac{2}{3}.\)
\(\displaystyle \int_{0}^{1} x^3\,dx
= \left[\frac{x^4}{4}\right]_{0}^{1}
= \frac{1}{4}.\)
Hence, the total area of \(S\) is
\[
\frac{2}{3} - \frac{1}{4}
= \frac{8}{12} - \frac{3}{12}
= \frac{5}{12}.
\]
Step 4: Subdivision of \(S\) by \(y = 2x\)
The line \(y = 2x\) only affects the portion of \(S\) when \(0 \le x \le \tfrac{1}{4}\). Beyond \(x=\tfrac{1}{4}\), we have \(2x > \sqrt{x}\), so the line is above the entire region from \(x=\tfrac{1}{4}\) to \(x=1\). Thus, two subregions arise for \(0 \le x \le \tfrac{1}{4}\):
Region below \(y = 2x\): bounded by \(y=x^3\) and \(y=2x\).
Region above \(y = 2x\): bounded by \(y=2x\) and \(y=\sqrt{x}\).
Step 5: Area Computations for \(0 \le x \le \tfrac{1}{4}\)
5.1: Area between \(y = x^3\) and \(y = \sqrt{x}\) from \(0\) to \(\tfrac{1}{4}\)
\[
A_{[0,\tfrac{1}{4}]}
= \int_{0}^{\tfrac{1}{4}} \bigl[\sqrt{x} - x^3\bigr] \,dx.
\]
Compute each part:
\[
\int_{0}^{\tfrac{1}{4}} \sqrt{x}\,dx
= \left[\frac{2}{3} x^{\tfrac{3}{2}}\right]_{0}^{\tfrac{1}{4}}
= \frac{2}{3}\left(\frac{1}{4}\right)^{\tfrac{3}{2}}
= \frac{2}{3}\cdot \frac{1}{8}
= \frac{1}{12}.
\]
\[
\int_{0}^{\tfrac{1}{4}} x^3\,dx
= \left[\frac{x^4}{4}\right]_{0}^{\tfrac{1}{4}}
= \frac{(1/4)^4}{4}
= \frac{1/256}{4}
= \frac{1}{1024}.
\]
Hence,
\[
A_{[0,\tfrac{1}{4}]}
= \frac{1}{12} - \frac{1}{1024}
= \frac{256}{3072} - \frac{3}{3072}
= \frac{253}{3072}.
\]
5.2: Area between \(y = x^3\) and \(y = 2x\) from \(0\) to \(\tfrac{1}{4}\)
\[
\int_{0}^{\tfrac{1}{4}} \bigl[2x - x^3\bigr] \,dx
= \int_{0}^{\tfrac{1}{4}} 2x \,dx
- \int_{0}^{\tfrac{1}{4}} x^3 \,dx.
\]
Compute separately:
\[
\int_{0}^{\tfrac{1}{4}} 2x \,dx
= \left[x^2\right]_{0}^{\tfrac{1}{4}}
= \left(\tfrac{1}{4}\right)^2
= \tfrac{1}{16},
\]
\[
\int_{0}^{\tfrac{1}{4}} x^3 \,dx
= \tfrac{1}{1024} \quad\text{(as above).}
\]
Thus,
\[
\int_{0}^{\tfrac{1}{4}} \bigl[2x - x^3\bigr] \,dx
= \frac{1}{16} - \frac{1}{1024}
= \frac{64}{1024} - \frac{1}{1024}
= \frac{63}{1024}.
\]
5.3: Area between \(y = 2x\) and \(y = \sqrt{x}\) from \(0\) to \(\tfrac{1}{4}\)
This is simply the difference between the total area from \(x^3\) to \(\sqrt{x}\) (step 5.1) and the area from \(x^3\) to \(2x\) (step 5.2) on \([\,0,\tfrac{1}{4}\,]\):
\[
\left(\frac{253}{3072}\right) - \left(\frac{63}{1024}\right).
\]
Since \(\frac{63}{1024} = \frac{189}{3072}\), the difference is
\[
\frac{253}{3072} - \frac{189}{3072} = \frac{64}{3072} = \frac{1}{48}.
\]
Step 6: Defining \(R_1\) and \(R_2\) and Calculating Their Areas
Because the problem states “If \(\max\{R_1, R_2\} = R_2\),” we infer that \(R_2\) must be the larger part. Let us identify:
\(R_2\) = the larger region,
\(R_1\) = the smaller region.
From our integrals, we see:
Above \(y = 2x\) on \([0, \tfrac{1}{4}]\): area = \(\tfrac{1}{48}\).
Below \(y = 2x\) on \([0, \tfrac{1}{4}]\): area = \(\tfrac{63}{1024}\).
From \(x=\tfrac{1}{4}\) to \(x=1\), the entire strip from \(y=x^3\) to \(y=\sqrt{x}\) is below \(y=2x\) (because \(2x\) is larger than \(\sqrt{x}\) there), so that entire area also belongs to the “below” region.
6.1: Area Below \(y=2x\)
\[
\text{Below Area}
= \int_{0}^{\tfrac{1}{4}} (2x - x^3)\,dx
\;+\;
\int_{\tfrac{1}{4}}^{1} \bigl[\sqrt{x} - x^3\bigr]\,dx.
\]
We already have the first part as \(\tfrac{63}{1024}\). The second part is
\[
\int_{\tfrac{1}{4}}^{1} \bigl[\sqrt{x} - x^3\bigr]\,dx
= \underbrace{\int_{0}^{1} \bigl[\sqrt{x}-x^3\bigr]\,dx}_{\tfrac{5}{12}}
- \underbrace{\int_{0}^{\tfrac{1}{4}} \bigl[\sqrt{x}-x^3\bigr]\,dx}_{\frac{253}{3072}}
= \frac{5}{12} - \frac{253}{3072}.
\]
Since \(\tfrac{5}{12} = \tfrac{1280}{3072}\), the difference is
\[
\frac{1280}{3072} - \frac{253}{3072} = \frac{1027}{3072}.
\]
Hence,
\[
\text{Below Area}
= \frac{63}{1024} \;+\; \frac{1027}{3072}.
\]
Converting \(\frac{63}{1024}\) to denominator \(3072\) gives \(\tfrac{189}{3072}\). Thus,
\[
\text{Below Area}
= \frac{189}{3072} + \frac{1027}{3072}
= \frac{1216}{3072}
= \frac{19}{48}.
\]
6.2: Area Above \(y=2x\)
This is only for \(0 \le x \le \tfrac{1}{4}\), from \(y = 2x\) to \(y = \sqrt{x}\), which we found to be \(\tfrac{1}{48}\).
Step 7: The Ratio \(\frac{R_2}{R_1}\)
We must match these with \(R_1\) and \(R_2\) such that \(R_2\) is the larger region. Observing the numerical values:
Area above \(y = 2x\) is \(\tfrac{1}{48}\)
Area below \(y = 2x\) is \(\tfrac{19}{48}\)
Clearly, \(\tfrac{19}{48}\) is bigger than \(\tfrac{1}{48}\). Thus:
\[
R_2 = \frac{19}{48},
\quad
R_1 = \frac{1}{48}.
\]
Hence the required ratio is
\[
\frac{R_2}{R_1}
= \frac{\tfrac{19}{48}}{\tfrac{1}{48}}
= 19.
\]
Final Answer
\(\displaystyle \frac{R_2}{R_1} = 19.\)
