The area (in sq. units) of the region described by {(x, y) : y² ≤ 2x, and y ≥ 4x - 1} is

Q: Why do we integrate with respect to $y$ instead of $x$?

Great question! If you integrate with respect to $x$, you have to split the region into two parts because the 'lower' boundary changes from the bottom half of the parabola to the line. By integrating with respect to $y$, the right boundary is always the line $x = \frac{y+1}{4}$ and the left boundary is always the parabola $x = \frac{y^2}{2}$. It simplifies the problem significantly and avoids potential errors in splitting the integral.

Q: How do we know which curve is the 'right' boundary?

Think about the horizontal position. For any given $y$-value within the region, the line $x = \frac{y+1}{4}$ is further to the right (has a larger $x$-value) than the parabola $x = \frac{y^2}{2}$. In the area formula $\int (x_{right} - x_{left}) dy$, we always subtract the left curve from the right curve to ensure the area is positive.

Q: I got confused with the negative limits. How do I handle them?

It is a common trap! When you substitute the lower limit $y = -\frac{1}{2}$ into the expression $\frac{y^2}{8} + \frac{y}{4} - \frac{y^3}{6}$, be extremely careful with the signs. Remember that $(-1/2)^2 = 1/4$ and $(-1/2)^3 = -1/8$. Writing each term out clearly before combining them is the best way to avoid a sign error.

Question

The area (in sq. units) of the region described by $\{(x, y) : y^2 \le 2x, 	ext{ and } y \ge 4x - 1\}$ is

Options:

(A) $\frac{11}{32}$

(B) $\frac{8}{9}$

(C) $\frac{11}{12}$

(D) $\frac{9}{32}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the Cartesian plane. You see a parabola, $y^2 = 2x$, opening its arms wide to the right, its vertex anchored at the origin.

Then, a line, $y = 4x - 1$, cuts across the scene. The region trapped between them is our target.

The inequality $y^2 \le 2x$ tells us we are inside the parabola, while $y \ge 4x - 1$ places us above or to the left of the line. This is the 'orange' slice of the plane we need to measure.

### The Strategic Choice: Why $dy$?

In many JEE problems, the choice of axis is the difference between a five-minute solution and a twenty-minute struggle. If we chose to integrate with respect to $x$, we would face a nightmare: the lower boundary changes, forcing us to split the integral into two pieces.

But look at the horizontal perspective! If we use horizontal strips, the right boundary is always the line:

$$x = \frac{y+1}{4}$$

The left boundary is always the parabola:

$$x = \frac{y^2}{2}$$

This is the path of least resistance.

### The Intersection: Finding the Limits

Before we can integrate, we need to know where our journey begins and ends. We equate the $x$-values of both curves:

$$\frac{y^2}{2} = \frac{y+1}{4}$$

Multiplying by $4$ gives us $2y^2 = y + 1$, or $2y^2 - y - 1 = 0$. Factoring this quadratic, we find $(2y + 1)(y - 1) = 0$.

Our intersection points are $y = 1$ and $y = -\frac{1}{2}$. These are our limits of integration.

### The Integration Masterclass

Now, we set up the integral:

$$A = \int_{-1/2}^{1} \left( \frac{y+1}{4} - \frac{y^2}{2} \right) dy$$

We integrate term by term: the integral of $\frac{y}{4}$ is $\frac{y^2}{8}$, the integral of $\frac{1}{4}$ is $\frac{y}{4}$, and the integral of $\frac{y^2}{2}$ is $\frac{y^3}{6}$. Our expression becomes:

$$\left[ \frac{y^2}{8} + \frac{y}{4} - \frac{y^3}{6} \right]_{-1/2}^{1}$$

### The Final Calculation

Substituting the upper limit $y = 1$ yields:

$$\frac{1}{8} + \frac{1}{4} - \frac{1}{6} = \frac{3+6-4}{24} = \frac{5}{24}$$

Now, for the lower limit $y = -\frac{1}{2}$, we must be vigilant with signs:

$$\frac{(-1/2)^2}{8} + \frac{-1/2}{4} - \frac{(-1/2)^3}{6} = \frac{1}{32} - \frac{1}{8} + \frac{1}{48}$$

Finding a common denominator of $96$, this becomes:

$$\frac{3 - 12 + 2}{96} = -\frac{7}{96}$$

Finally, the area is the upper limit value minus the lower limit value:

$$\frac{5}{24} - \left( -\frac{7}{96} \right) = \frac{20}{96} + \frac{7}{96} = \frac{27}{96}$$

Simplifying by dividing both numerator and denominator by $3$, we arrive at our final answer: **$\frac{9}{32}$ square units**.

Animated Solution for Mathematics - Definite Integration: The area (in sq. units) of the region described by ${(x, y) : y^{2} \leq 2 x, and y \geq 4 x - 1}$ is

Select Answer:

Visualized Solution

The Parabola $y^{2} = 2 x$

The Line $y = 4 x - 1$

Identifying the Region

Finding Intersection Points

Solving the Quadratic

Intersection Limits

Choosing the Integration Axis

Setting up the Integral

Integrating the Expression

Upper Limit Substitution

Lower Limit Substitution

Simplifying the Lower Limit

Final Area Calculation

Final Answer

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Strategic Choice

The Intersection

The Integration Masterclass

The Final Calculation

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area (in sq. units) of the region described by {(x,y):y2≤2x, and y≥4x−1} is