The area of the region enclosed by the parabola y = 4x - x² and 3y = (x - 4)² is equal to (1) 32/9 (

Q: Why do we subtract the curves? Can't we just integrate them separately?

Think of the integral as a 'net' area. If you integrate $y_1$ and $y_2$ separately, you are calculating the area under each curve down to the $x$-axis. By subtracting them, you are effectively 'cutting out' the empty space between the lower curve and the $x$-axis, leaving only the area trapped between the two curves. It is the most elegant way to isolate the region of interest.

Q: How do I know which curve is the upper one without drawing a graph?

You can test a point! Pick any value between your intersection points, say $x = 2$. Plug $x = 2$ into both equations. For $y = 4x - x^2$, you get $y = 8 - 4 = 4$. For $3y = (x - 4)^2$, you get $3y = (-2)^2 = 4$, so $y = 4/3$. Since $4 > 4/3$, the first curve is clearly the upper one in that interval. Testing a point is a foolproof way to verify your setup.

Q: Is there a shortcut formula for the area between two parabolas?

Yes, for two parabolas of the form $y = a_1x^2 + b_1x + c_1$ and $y = a_2x^2 + b_2x + c_2$, the area is $\frac{|a_1 - a_2|}{6} (\beta - \alpha)^3$, where $\alpha$ and $\beta$ are the roots. While this is a powerful tool for competitive exams, I encourage you to master the integration method first. Understanding the 'why' is what makes you a physicist, not just a calculator.

Question

The area of the region enclosed by the parabola $y = 4x - x^2$ and $3y = (x - 4)^2$ is equal to

Options:

(A) 32/9

(B) 4

(C) 6

(D) 14/3

Sigma Percentile · Accepted Answer

### The Dance of the Parabolas: Unveiling the Hidden Area

Welcome, fellow traveler on the path of JEE mastery. Today, we are not just solving a math problem; we are witnessing a geometric dance.

We have two parabolas, $y = 4x - x^2$ and $3y = (x - 4)^2$, and they have conspired to trap a region of space between them. Our mission is to measure that space. This is a classic problem that tests your ability to visualize, set up, and execute with precision.

### Phase 1: The Visualization

Before we touch a pen to paper, let us look at the curves. The first, $y = 4x - x^2$, is a classic downward-opening parabola. The negative coefficient of $x^2$ tells us it is frowning at the origin.

The second, $3y = (x - 4)^2$, is an upward-opening parabola, smiling at the sky. When you place these two on a coordinate plane, they intersect at two distinct points, creating a closed, eye-shaped region. This is the area we need to calculate.

### Phase 2: Finding the Boundaries

To measure this area, we need to know where it begins and where it ends. These are our limits of integration. At the points of intersection, the $y$-values of both curves must be identical.

So, we set them equal:

$$4x - x^2 = \frac{(x - 4)^2}{3}$$

This is where the algebra begins. We multiply by $3$ to clear the fraction, giving us $12x - 3x^2 = (x - 4)^2$. Expanding the right side, we get $12x - 3x^2 = x^2 - 8x + 16$.

Bringing all terms to one side, we arrive at $4x^2 - 20x + 16 = 0$. Dividing by $4$, we find the beautiful, simple quadratic: $x^2 - 5x + 4 = 0$. Factoring this, we get $(x - 1)(x - 4) = 0$. Our boundaries are set: $x = 1$ and $x = 4$.

### Phase 3: The Integral Setup

Now, we invoke the power of calculus. The area $A$ between two curves from $x = a$ to $x = b$ is the integral of the upper curve minus the lower curve. We have already determined that $y = 4x - x^2$ is the upper curve in this interval.

Thus, our integral is:

$$A = \int_{1}^{4} \left[ (4x - x^2) - \frac{(x - 4)^2}{3} \right] \, dx$$

This looks intimidating, but let us break it down. We integrate term by term. The integral of $4x$ is $2x^2$. The integral of $x^2$ is $\frac{x^3}{3}$. For the term $\frac{(x - 4)^2}{3}$, we use the power rule to get $\frac{(x - 4)^3}{9}$.

### Phase 4: The Final Calculation

Our antiderivative is:

$$\left[ 2x^2 - \frac{x^3}{3} - \frac{(x - 4)^3}{9} \right]_{1}^{4}$$

First, we evaluate at the upper limit, $x = 4$:

$$2(4)^2 - \frac{4^3}{3} - \frac{(4 - 4)^3}{9} = 32 - \frac{64}{3} - 0 = \frac{32}{3}$$

Next, we evaluate at the lower limit, $x = 1$:

$$2(1)^2 - \frac{1^3}{3} - \frac{(1 - 4)^3}{9} = 2 - \frac{1}{3} - \frac{-27}{9} = 2 - \frac{1}{3} + 3 = 5 - \frac{1}{3} = \frac{14}{3}$$

Finally, we subtract the lower limit value from the upper limit value:

$$A = \frac{32}{3} - \frac{14}{3} = \frac{18}{3} = 6$$

The area is exactly **$6$ square units**. It is a clean, elegant result. Remember, in JEE, the math is often just the language; the real skill is in the visualization and the careful, step-by-step execution. Keep practicing, keep visualizing, and you will master these curves.

Animated Solution for Mathematics - Definite Integration: The area of the region enclosed by the parabola $y = 4 x - x^{2}$ and $3 y = (x - 4)^{2}$ is equal to

Select Answer:

Visualized Solution

Visualize the Curves

Find Intersection Points

Equate the Equations

Expand and Rearrange

Solve the Quadratic Equation

Identify the Limits

Set Up the Area Integral

Integrate the Expression

Apply the Upper Limit

Apply the Lower Limit

Calculate Final Area

The Sigma Insight: Area Bounded by Curves

The Dance of the Parabolas

Phase 1

Phase 2

Phase 3

Phase 4

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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 of the region enclosed by the parabola y=4x−x2 and 3y=(x−4)2 is equal to