The area (in sq. units) of the region described by {(x, y) : y² ≤ 2x and y ≥ 4x-1} is
(a) 15/64
(b)

Q: Why can't we just integrate with respect to $x$? It feels more natural to me.

That is a fantastic question! You absolutely can integrate with respect to $x$, but look at the geometry. If you integrate w.r.t $x$, you have to deal with the square root $\sqrt{2x}$ and split the integral at the vertex of the parabola. Integrating w.r.t $y$ turns a complex problem into a simple polynomial integral. In JEE, choosing the right variable is half the battle—always look for the path of least resistance!

Q: How do I know which function is the 'right' one in the integral?

Always sketch the graph! The 'right' curve is the one with larger $x$-values for a given $y$. In our case, the line is to the right of the parabola within the intersection interval. If you are ever unsure, just pick a test point between the intersection values and see which $x$ is larger. The curve with the larger $x$ value is your upper boundary for the integral.

Q: Why did the integral result in a negative value for the lower limit calculation?

Don't panic! That is perfectly normal. When you evaluate a definite integral, the antiderivative can take on negative values depending on where the curve lies relative to the axes. The final area is the difference between the upper limit evaluation and the lower limit evaluation. The negative sign is just part of the arithmetic process, and it will resolve itself when you perform the final subtraction.

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) 15/64

(B) 9/32

(C) 7/32

(D) 5/64

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane. You have two entities: a parabola, $y^2 = 2x$, which is a smooth, sweeping curve opening to the right, and a line, $y = 4x - 1$, which cuts through the plane with linear precision.

Our goal is to find the area of the region trapped between them. This is not just a calculation; it is a dance between two functions. To solve this, we must first understand the boundaries of this dance.

### Phase 1: The Intersection

Before we can measure the area, we must know where the dance begins and ends. We need the intersection points.

If we express $x$ in terms of $y$, the parabola becomes $x = \frac{y^2}{2}$ and the line becomes $x = \frac{y+1}{4}$. By equating these, we get:

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

Multiplying by $4$, we arrive at $2y^2 = y + 1$, or the quadratic equation:

$$ 2y^2 - y - 1 = 0 $$

Factoring this gives us $(2y+1)(y-1) = 0$. The intersection points are at $y = 1$ and $y = -\frac{1}{2}$. This is the beauty of choosing the right variable—the algebra becomes elegant and simple.

### Phase 2: The Calculus of Area

Now that we have our limits, we set up the integral. The area is the integral of the 'right' curve minus the 'left' curve.

Since the line is to the right of the parabola in this region, our integral is:

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

This is a simple polynomial integral. 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 $-\frac{y^3}{6}$.

We have our antiderivative:

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

### Phase 3: The Final Tally

Now, we evaluate the expression. At the upper limit $y = 1$, we get:

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

At the lower limit $y = -\frac{1}{2}$, we get:

$$ \frac{1}{32} - \frac{1}{8} + \frac{1}{48} = -\frac{7}{96} $$

Subtracting the lower from the upper, we get:

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

Simplifying this fraction, we arrive at our final answer: **$\frac{9}{32}$ square units**. You have successfully navigated the geometry, the algebra, and the calculus.

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

Visualizing the Region

Finding Intersection Points

Solving the Quadratic Equation

Determining the Coordinates

Setting up the Area Integral

Integrating Term by Term

Evaluating at Upper Limit $y = 1$

Evaluating at Lower Limit $y = - \frac{1}{2}$

Final Area Calculation

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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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