If the area of the region bounded by the curves y = 4 - x²/4 and y = (x-4)/2 is equal to α, then 6α

Q: Why is the parabola the upper curve? How do I know for sure without graphing it perfectly?

That is a brilliant question. In the interval $[-6, 4]$, pick a test point, say $x=0$. The parabola gives $y=4$, while the line gives $y=-2$. Since $4 > -2$, the parabola is definitely on top. Always use a test point if you are unsure about the relative positions of the curves.

Q: I keep messing up the signs when evaluating the lower limit. Any tips?

The negative sign is the silent killer in JEE. When evaluating $\int_{a}^{b} f(x) dx = F(b) - F(a)$, treat $F(a)$ as a separate entity inside a bracket. Always write $F(b) - (F(a))$ to avoid distribution errors. When you have a negative limit like $-6$, the powers of $x$ can flip signs unexpectedly, so be extremely deliberate.

Q: Why did the question ask for $6\alpha$ instead of just $\alpha$?

This is a classic JEE Advanced tactic. They want to test if you are reading the question carefully until the very last second. It is a test of composure. Never celebrate too early; always re-read the final line of the question before bubbling your answer to ensure you have performed the final operation required.

Question

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 $\alpha$, then $6\alpha$ equals

Options:

(A) 250

(B) 210

(C) 240

(D) 220

Sigma Percentile · Accepted Answer

### Analyzing the Geometry

Welcome, future engineer. Today, we are not just solving an integral; we are exploring the space between two mathematical entities. We have a downward-opening parabola, $y = 4 - \frac{x^2}{4}$, and a straight line, $y = \frac{x-4}{2}$.

Imagine these two curves on a coordinate plane. The parabola is a hill, and the line is a path cutting through it. The region bounded by them is the area we need to calculate.

To find this area, we first need to know where these two paths meet. We set their $y$-values equal:

$$4 - \frac{x^2}{4} = \frac{x-4}{2}$$

### The Algebra of Intersection

To solve this, let us clear the denominators by multiplying the entire equation by $4$. This yields $16 - x^2 = 2(x-4)$, which simplifies to $16 - x^2 = 2x - 8$.

Bringing everything to one side, we get the quadratic equation:

$$x^2 + 2x - 24 = 0$$

Factoring this is straightforward: $(x+6)(x-4) = 0$. Thus, our intersection points are $x = -6$ and $x = 4$. These are the boundaries of our integral.

### The Integral Setup

Now, we define our area $\alpha$ as the integral of the upper curve minus the lower curve. As we verified with a test point, the parabola is the upper boundary.

So, the integral is defined as:

$$\alpha = \int_{-6}^{4} \left( \left( 4 - \frac{x^2}{4} ight) - \left( \frac{x-4}{2} ight) ight) dx$$

Before we integrate, let us simplify the integrand. Distributing the negative sign and combining the constants, we arrive at:

$$\alpha = \int_{-6}^{4} \left( 6 - \frac{x}{2} - \frac{x^2}{4} ight) dx$$

### The Integration Grind

Now, we integrate term by term. The integral of $6$ is $6x$, the integral of $-\frac{x}{2}$ is $-\frac{x^2}{4}$, and the integral of $-\frac{x^2}{4}$ is $-\frac{x^3}{12}$.

Our expression becomes:

$$\left[ 6x - \frac{x^2}{4} - \frac{x^3}{12} ight]_{-6}^{4}$$

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

$$6(4) - \frac{4^2}{4} - \frac{4^3}{12} = 24 - 4 - \frac{64}{12} = 20 - \frac{16}{3} = \frac{44}{3}$$

Next, we evaluate at the lower limit $x = -6$:

$$6(-6) - \frac{(-6)^2}{4} - \frac{(-6)^3}{12} = -36 - \frac{36}{4} - \frac{-216}{12} = -36 - 9 + 18 = -27$$

### Final Calculation

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

$$\alpha = \frac{44}{3} - (-27) = \frac{44}{3} + 27 = \frac{44 + 81}{3} = \frac{125}{3}$$

The question asks for $6\alpha$. We multiply our result by $6$:

$$6 	imes \frac{125}{3} = 2 	imes 125 = 250$$

The final result is **250**. You have successfully navigated the geometry, the algebra, and the calculus. Keep this focus, and no problem will be too difficult for you.

Animated Solution for Mathematics - Definite Integration: 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

Select Answer:

Visualized Solution

Identify the Curves

Setting up Intersection

Solving for $x$

Finding the Roots

The Bounded Region

Setting up the Integral

Simplifying the Integrand

Integration Process

Evaluating Upper Limit

Evaluating Lower Limit

Calculating $α$

Final Answer

The Sigma Insight: Area Bounded by Curves

Analyzing the Geometry

The Algebra of Intersection

The Integral Setup

The Integration Grind

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; } If the area of the region bounded by the curves y=4−4x2​ and y=2x−4​ is equal to α, then 6α equals