Let the area of the region {(x,y) : |2x-1| ≤ y ≤ |x²-x|, 0 ≤ x ≤ 1} be A. Then (6A+11)² is equal to

Q: Why did we replace $|x^2-x|$ with $x-x^2$?

Great observation! In the interval $[0, 1]$, the parabola $f(x) = x^2-x$ opens upwards and its roots are $0$ and $1$. Between these roots, the function values are negative. Since the absolute value $|f(x)|$ must be non-negative, we flip the sign: $|x^2-x| = -(x^2-x) = x-x^2$.

Q: How do I know which function is the upper boundary?

Always test a point! Pick a value in your interval, say $x = 0.6$. At $x = 0.6$, $y = |2(0.6)-1| = 0.2$, while $y = |0.6^2-0.6| = 0.24$. Since $0.24 > 0.2$, the parabola is the upper boundary in that specific sub-interval. Visualizing the graph is the most reliable way to avoid sign errors.

Q: Is it mandatory to use symmetry, or can I just integrate from the first intersection to the last?

You can absolutely integrate across the whole range! However, symmetry is a powerful weapon for a JEE aspirant. It reduces the chance of arithmetic errors and saves precious time. If you see a mirror image, use it to your advantage.

Q: The quadratic roots look messy. Did I make a mistake?

Not at all! In JEE Advanced, you will often encounter irrational roots involving $\sqrt{5}$. The key is to keep them in their radical form as long as possible. Notice how the term $\alpha^2 + \alpha - 1 = 0$ allowed us to simplify the cubic terms later. Trust the process!

Q: Why does the final expression $(6A+11)^2$ look so clean?

That is the beauty of well-designed problems! The complexity of the integral is designed to 'cancel out' or simplify into a form that yields a perfect integer. If your intermediate steps look like a nightmare, take a breath—often, a clever substitution or algebraic identity is waiting to collapse the complexity.

Question

Let the area of the region $\{(x,y) : |2x-1| \le y \le |x^2-x|, 0 \le x \le 1\}$ be $A$. Then $(6A+11)^2$ is equal to ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE excellence. Today, we are tasked with finding the area $A$ of a region defined by two absolute value functions: $y = |2x-1|$ and $y = |x^2-x|$ within the domain $x \in [0, 1]$.

In the interval $[0, 1]$, the expression $x^2-x$ is always non-positive. Therefore, the absolute value $|x^2-x|$ simplifies to $x-x^2$.

We now have two clear functions: $y = |2x-1|$, which is a V-shaped graph with a vertex at $x = 1/2$, and $y = x-x^2$, which is an inverted parabola.

### Visualizing the Symmetry

These two shapes intersect, creating a trapped region. The symmetry here is striking—the entire region is symmetric about the vertical line $x = 1/2$.

This is our first "gift" from the problem. It means we only need to calculate half the area and double it to find the total area $A$.

### The Hunt for Intersections

To find the boundaries of our integral, we set $|2x-1| = x-x^2$. Because of the absolute value, we split this into two cases.

For $x \ge 1/2$, the equation becomes $2x-1 = x-x^2$, which simplifies to the quadratic equation:

$$x^2 + x - 1 = 0$$

Using the quadratic formula, we find the root in our interval:

$$\alpha = \frac{\sqrt{5}-1}{2}$$

This value $\alpha$ serves as our upper limit of integration for the right half of the region.

### The Calculus of Symmetry

Given the symmetry identified earlier, the total area $A$ is defined by the following integral:

$$A = 2 \int_{\frac{1}{2}}^{\frac{\sqrt{5}-1}{2}} ((x-x^2) - (2x-1)) dx$$

Simplifying the integrand, we obtain $-x^2 - x + 1$. Integrating this polynomial term by term yields:

$$\int (-x^2 - x + 1) dx = -\frac{x^3}{3} - \frac{x^2}{2} + x$$

### Final Calculation

Evaluating this at the irrational limit $\alpha = \frac{\sqrt{5}-1}{2}$ is simplified by the identity $\alpha^2 = 1-\alpha$. By substituting this into our evaluated expression, the terms collapse into a much simpler form.

After evaluating the limits, we find the area:

$$A = \frac{5\sqrt{5}-11}{6}$$

We are asked to compute $(6A+11)^2$. Multiplying our area $A$ by $6$ yields $5\sqrt{5}-11$. Adding $11$ leaves us with $5\sqrt{5}$.

Squaring this result provides the final, satisfying answer:

$$(5\sqrt{5})^2 = 25 	imes 5 = \mathbf{125}$$

Animated Solution for Mathematics - Definite Integration: Let the area of the region ${(x, y) : ∣2 x - 1∣ \leq y \leq ∣ x^{2} - x ∣, 0 \leq x \leq 1}$ be $A$ . Then $(6 A + 11)^{2}$ is equal to ______.

Enter Numerical Value:

Visualized Solution

Visualizing the Region

Finding Intersection Points

Solving Case 1: $x \geq \frac{1}{2}$

Solving Case 2: $x < \frac{1}{2}$

Exploiting Symmetry

Setting up the Integral

Evaluating at Upper Limit

Evaluating at Lower Limit

Calculating Total Area $A$

Final Computation

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Visualizing the Symmetry

The Hunt for Intersections

The Calculus of Symmetry

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; } Let the area of the region {(x,y):∣2x−1∣≤y≤∣x2−x∣,0≤x≤1} be A. Then (6A+11)2 is equal to ______.