The area (in sq. units) of the region {(x, y) ∈ R² | 4x² ≤ y ≤ 8x + 12} is:
(A) 127/3
(B) 128/3
(C)

Q: Why do we subtract the parabola from the line? Can't we do it the other way around?

That is a fantastic question! The area of a region must always be a positive quantity. In the interval between our intersection points, the line $y = 8x + 12$ lies above the parabola $y = 4x^2$. By calculating $y_{upper} - y_{lower}$, we ensure the integrand is positive, yielding a positive area. If you reversed it, you would get the negative of the area. Think of it as measuring the 'height' of the vertical strips—height must be positive!

Q: Is it possible to solve this using geometry instead of calculus?

You have a sharp eye! Yes, there is a specific property for the area of a parabolic segment (Archimedes' Quadrature of the Parabola), which states the area is $\frac{4}{3}$ of the area of the triangle formed by the chord and the tangent parallel to it. However, for JEE Advanced, mastering the calculus approach is non-negotiable because it is the universal tool that works even when the curves are not simple parabolas. Stick to the integral method—it is the most robust path.

Q: I keep getting a negative answer when evaluating the integral. What am I doing wrong?

Don't panic! This is a classic 'sign trap.' Usually, this happens because of a mistake in the evaluation of the lower limit. Remember, the Fundamental Theorem of Calculus is $F(b) - F(a)$. If $a$ is negative, you are subtracting a negative value, which becomes addition. Double-check your arithmetic when plugging in $x = -1$. It is very easy to lose a negative sign when cubing or squaring negative numbers!

Question

The area (in sq. units) of the region $\{(x, y) \in R^2 | 4x^2 \le y \le 8x + 12\}$ is:

Options:

(A) $\frac{127}{3}$

(B) $\frac{128}{3}$

(C) $\frac{124}{3}$

(D) $\frac{125}{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in front of a vast, empty coordinate plane. You have two distinct entities: a parabola, $y = 4x^2$, which is a gentle, upward-opening curve, and a line, $y = 8x + 12$, which slices across the plane with a steady, relentless slope.

The problem asks us to find the area of the region trapped between them. This is not just an algebraic exercise; it is a geometric dance where the parabola defines the floor and the line defines the ceiling.

Before we touch a single variable, we must visualize the intersection. Defining the boundaries is the first step in any JEE problem.

### The Intersection: The Moment of Truth

To find the limits of our integration, we must find the points where these two curves meet. At the intersection, the $y$-values are identical.

Thus, we equate them:
$$4x^2 = 8x + 12$$

We bring all terms to one side to form a standard quadratic equation:
$$4x^2 - 8x - 12 = 0$$

The coefficients are all divisible by $4$. Dividing by $4$, we get:
$$x^2 - 2x - 3 = 0$$

Factoring this is straightforward:
$$(x - 3)(x + 1) = 0$$

This gives us our limits: $x = -1$ and $x = 3$. These are the vertical walls of our room, meaning we are integrating from $x = -1$ to $x = 3$.

### The Calculus: The Art of Summation

Now, we enter the heart of the problem: the definite integral. The area $A$ is the accumulation of infinitely thin vertical strips, each with a height equal to the difference between the upper curve and the lower curve.

Mathematically, this is expressed as:
$$A = \int_{-1}^{3} (y_{upper} - y_{lower}) dx$$

Substituting our functions, we get:
$$A = \int_{-1}^{3} (8x + 12 - 4x^2) dx$$

We integrate term-by-term using the power rule, $\int x^n dx = \frac{x^{n+1}}{n+1}$. The integral becomes:
$$A = \left[ 4x^2 + 12x - \frac{4x^3}{3} \right]_{-1}^{3}$$

### The Final Tally: Precision and Patience

We apply the Fundamental Theorem of Calculus: $F(3) - F(-1)$. First, we evaluate at the upper limit, $x = 3$:
$$4(3)^2 + 12(3) - \frac{4(3)^3}{3} = 36 + 36 - 36 = 36$$

Next, we evaluate at the lower limit, $x = -1$:
$$4(-1)^2 + 12(-1) - \frac{4(-1)^3}{3} = 4 - 12 + \frac{4}{3} = -8 + \frac{4}{3} = -\frac{20}{3}$$

Finally, we subtract the lower limit value from the upper limit value:
$$A = 36 - \left( -\frac{20}{3} \right) = 36 + \frac{20}{3}$$

Converting to a common denominator, we get:
$$\frac{108 + 20}{3} = \frac{128}{3}$$

The final area is **$\frac{128}{3}$ square units**. You have mastered the geometry, the algebra, and the calculus required to solve this problem.

Animated Solution for Mathematics - Definite Integration: The area (in sq. units) of the region ${(x, y) \in R^{2} ∣4 x^{2} \leq y \leq 8 x + 12}$ is:

Select Answer:

Visualized Solution

Defining the Region

Identifying the Bounded Area

Equating the Boundaries

Forming the Quadratic Equation

Finding the Limits of Integration

The Area Formula

Substituting the Curves

Performing the Integration

Evaluating at $x = 3$

Evaluating at $x = - 1$

Calculating the Total Area

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Intersection

The Calculus

The Final Tally

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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 {(x,y)∈R2∣4x2≤y≤8x+12} is: