Let S be the region bounded by the curves y = x³ and y² = x. The curve y = 2|x| divides S into two r

Q: Why can we just ignore the absolute value in $y=2|x|$?

Great observation! The absolute value function $|x|$ behaves differently for negative and positive $x$. However, look at our region $S$. It is bounded by $y=x^3$ and $y^2=x$. Since $y^2=x$ implies $x \ge 0$, our entire region $S$ exists strictly in the first quadrant. Thus, for all points in $S$, $x$ is non-negative, and $|x| = x$. We are essentially working in a world where $x$ is always positive.

Q: Is it possible to integrate with respect to $y$ instead of $x$?

Absolutely! You could express $x$ as a function of $y$ (i.e., $x=y^2$ and $x=y^{1/3}$) and integrate with respect to $y$. It is a fantastic exercise to verify your answer using this method. However, integrating with respect to $x$ is often more intuitive when the functions are already given in the form $y=f(x)$. Both paths lead to the same mountain peak!

Q: How do I know which curve is the 'upper' curve?

Always test a point! Pick a value between your intersection points, say $x=0.25$. Calculate the $y$-values for both curves at this point. The curve with the larger $y$-value is your 'upper' curve for that interval. Visualizing the graph is the most powerful tool in your arsenal, but testing a point is the ultimate safety net.

Question

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 ____.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are mapping a territory. We are looking at a region $S$ defined by two curves, $y = x^3$ and $y^2 = x$.

This is a classic JEE Advanced scenario where the beauty lies not in the complexity of the calculation, but in the elegance of the visualization. Imagine standing on a coordinate plane where the cubic curve $y = x^3$ rises steadily and the parabola $y^2 = x$ (or $y = \sqrt{x}$ in the first quadrant) arches gracefully. They meet and enclose a space, which is our region $S$.

### Defining the Boundaries

Before we touch a single integral, we must determine the intersection points. We set $x^3 = \sqrt{x}$, which leads to the equation:

$$x^6 = x \Rightarrow x(x^5 - 1) = 0$$

The roots are $x=0$ and $x=1$. This confirms that our region $S$ is confined between $x=0$ and $x=1$.

Within this interval, we test $x = 0.5$ to identify the upper boundary. Since $\sqrt{0.5} \approx 0.707$ and $(0.5)^3 = 0.125$, the square root function is the ceiling and the cubic function is the floor. The total area $S$ is given by:

$$S = \int_{0}^{1} (\sqrt{x} - x^3) dx$$

Evaluating this integral, we find:

$$S = \left[ \frac{2}{3}x^{3/2} - \frac{x^4}{4} \right]_0^1 = \frac{2}{3} - \frac{1}{4} = \frac{5}{12}$$

### The Intruder

The problem introduces a divider: $y = 2|x|$. In the first quadrant, this simplifies to the line $y = 2x$. This line starts at the origin and slices through the upper boundary of our region.

To find where it exits, we set $2x = \sqrt{x}$. Squaring both sides yields $4x^2 = x$, which results in $x = 1/4$. This line partitions our region $S$ into two parts: $R_1$ and $R_2$.

### The Calculation of $R_1$

We focus on $R_1$, the smaller region trapped between the line $y = 2x$ and the upper curve $y = \sqrt{x}$ from $x=0$ to $x=1/4$. The integral is:

$$R_1 = \int_{0}^{1/4} (\sqrt{x} - 2x) dx$$

Integrating term by term, we obtain:

$$R_1 = \left[ \frac{2}{3}x^{3/2} - x^2 \right]_0^{1/4}$$

Substituting the upper limit $x = 1/4$:

$$R_1 = \frac{2}{3}\left(\frac{1}{4}\right)^{3/2} - \left(\frac{1}{4}\right)^2 = \frac{2}{3}\left(\frac{1}{8}\right) - \frac{1}{16} = \frac{1}{12} - \frac{1}{16}$$

Calculating the difference, we get:

$$R_1 = \frac{4 - 3}{48} = \frac{1}{48}$$

### Final Calculation

We know the total area $S = 5/12$ and the smaller area $R_1 = 1/48$. The remaining area $R_2$ is $S - R_1$. Converting $S$ to a denominator of 48, we have $S = 20/48$.

Thus, the area of the second region is:

$$R_2 = \frac{20}{48} - \frac{1}{48} = \frac{19}{48}$$

The problem asks for the ratio $R_2 / R_1$. Calculating this:

$$\frac{R_2}{R_1} = \frac{19/48}{1/48} = 19$$

The final ratio is **19**.

Animated Solution for Mathematics - Definite Integration: 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 ____.

Enter Numerical Value:

Visualized Solution

Visualizing the Curves $y = x^{3}$ and $y^{2} = x$

Finding Intersection Points

Identifying Region $S$

Setting up the Integral for $S$

Evaluating the Area $S$

Introducing the Divider $y = 2∣ x ∣$

Intersection of Divider and $y = x$

Identifying Region $R_{1}$

Setting up Integral for $R_{1}$

Computing Area $R_{1}$

Calculating Area $R_{2}$

Final Ratio $\frac{R _{2}}{R _{1}}$

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Defining the Boundaries

The Intruder

The Calculation of $R_{1}$

Final Calculation

Similar Questions

The area of the region bounded by the curve $y = max {∣ x ∣, x ∣ x - 2∣}$ , then x-axis and the lines $x = - 2$ and $x = 4$ is equal to _______ .

The area enclosed between the curves $y^{2} = x$ and $y = ∣ x ∣$ is

The area enclosed between the curves $y = x ∣ x ∣$ and $y = x - ∣ x ∣$ is :

Find the area of the region bounded by the curves $y = x^{2}, y = 2 - x^{2}$ and $y = 2$ , which lies to the right of the line $x = 1$ .

Let $α$ be the area of the larger region bounded by the curve $y^{2} = 8 x$ and the lines $y = x$ and $x = 2$ , which lies in the first quadrant. Then the value of $3 α$ is equal to ______.

The area bounded by the curves $y = ∣ x^{2} - 1∣$ and $y = 1$ is

Let $f (x) =$ Maximum ${x^{2}, (1 - x)^{2}, 2 x (1 - x)}$ , where $0 \leq x \leq 1$ . Determine the area of the region bounded by the curves $y = f (x), x$ -axis, $x = 0$ and $x = 1$ .

Sketch the region bounded by the curves $y = x^{2}$ and $y = \frac{2}{1 + x ^{2}}$ . Find the area.

If the area of the region bounded by the curves $y = 4 - \frac{x ^{2}}{4}$ and $y = \frac{x - 4}{2}$ is equal to $α$ , then $6 α$ equals

Let for $x \in R$ , $f (x) = \frac{x + ∣ x ∣}{2}$ and $g (x) = {x, x^{2}, x < 0 x \geq 0$ . Then area bounded by the curve $y = (f \circ g) (x)$ and the lines $y = 0, 2 y - x = 15$ is equal to ______.

Enter Numerical Value:

Visualized Solution

Visualizing the Curves .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=x3 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y2=x

Finding Intersection Points

Identifying Region .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Setting up the Integral for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Evaluating the Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Introducing the Divider .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=2∣x∣

Intersection of Divider and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=x​

Identifying Region .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​

Setting up Integral for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​

Computing Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​

Calculating Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R2​

Final Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​R2​​