Given : f(x) = { x, 0 ≤ x

Q: Wait, why did we ignore the first two parts of the piecewise function $f(x)$?

Great observation! In calculus, we only care about the behavior of the function within the bounds of integration. Since our interval is $[\frac{1}{2}, \frac{\sqrt{3}}{2}]$, we only need the definition of $f(x)$ that applies to this specific range. The rest of the function is irrelevant to the area calculation.

Q: How do I know which function is the 'upper' curve without graphing it?

You can pick a test point inside the interval, say $x = 0.6$. Calculate $f(0.6) = 1 - 0.6 = 0.4$ and $g(0.6) = (0.6 - 0.5)^2 = 0.01$. Since $0.4 > 0.01$, $f(x)$ is clearly the upper curve. Always verify this to ensure your area result remains positive!

Q: Is it possible for the curves to cross each other inside the interval?

Excellent question. If they crossed, the area would be split into two integrals. However, by checking the endpoints, we saw they intersect exactly at $x = \frac{\sqrt{3}}{2}$. Since they don't cross within the interior of the interval, we can safely integrate the difference as a single expression.

Q: Why did the linear terms in the integrand cancel out?

That is the beauty of the algebra here! When you expand $(x - \frac{1}{2})^2$, you get $x^2 - x + \frac{1}{4}$. Subtracting this from $(1 - x)$ results in $(1 - x) - (x^2 - x + \frac{1}{4}) = 1 - x - x^2 + x - \frac{1}{4}$. The $-x$ and $+x$ terms vanish, leaving a much simpler quadratic. It's a sign that the problem is designed with mathematical elegance in mind.

Q: What if I made a mistake in the lower limit calculation?

Arithmetic errors are the most common trap in JEE Advanced. Always write out your fractions clearly. For the lower limit $x = \frac{1}{2}$, we have $\frac{3}{4}(\frac{1}{2}) - \frac{1}{3}(\frac{1}{8}) = \frac{3}{8} - \frac{1}{24}$. Finding a common denominator of 24 gives $\frac{9-1}{24} = \frac{8}{24} = \frac{1}{3}$. Take your time with these steps!

Question

Given : $f(x) = \begin{cases} x, & 0 \le x < \frac{1}{2} \ \frac{1}{2}, & x = \frac{1}{2} \ 1-x, & \frac{1}{2} < x \le 1 \end{cases}$ and $g(x) = (x - \frac{1}{2})^2, x \in R$. Then the area (in sq. units) of the region bounded by the curves, $y = f(x)$ and $y = g(x)$ between the line, $2x = 1$ and $2x = \sqrt{3}$, is:

Options:

(A) $\frac{\sqrt{3}}{4} - \frac{1}{3}$

(B) $\frac{1}{2} + \frac{\sqrt{3}}{4}$

(C) $\frac{1}{2} + \frac{\sqrt{3}}{4} - \frac{1}{3}$

(D) $4\sqrt{3} - \frac{1}{3}$

Sigma Percentile · Accepted Answer

### The Geometry of Hidden Elegance

Welcome, fellow traveler on the JEE journey. Today, we are not just solving an integral; we are peeling back the layers of a function to reveal a beautiful geometric truth.

When you first look at a piecewise function like $f(x)$, it is natural to feel a moment of hesitation. It looks like a fragmented puzzle, doesn't it? But remember, in the world of calculus, we are the masters of the domain. We decide where to look, and we decide what matters.

### Phase 1: Defining Our Territory

The problem asks for the area between $2x = 1$ and $2x = \sqrt{3}$. Before we touch a single integral sign, let us translate these boundaries into the language of $x$.

Dividing by 2, we find our interval is $[\frac{1}{2}, \frac{\sqrt{3}}{2}]$.

Now, look at the definition of $f(x)$. It changes its behavior at $x = \frac{1}{2}$. Since our interval starts exactly at $\frac{1}{2}$ and moves to the right, we only care about the branch $f(x) = 1 - x$.

The rest of the function exists, but it is not part of our story today. We discard the noise and focus on the signal.

### Phase 2: The Dance of the Curves

We have two players: $f(x) = 1 - x$ and $g(x) = (x - \frac{1}{2})^2$. To find the area between them, we need to know who is on top.

We check the boundaries. At $x = \frac{\sqrt{3}}{2}$, both functions yield $1 - \frac{\sqrt{3}}{2}$. They meet at the finish line!

This is a wonderful sign. It means our region is neatly enclosed without any messy crossings. By testing a point, we confirm that $f(x) \ge g(x)$ throughout our interval. We are ready to set up the integral:

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

### Phase 3: The Algebraic Symphony

Now, let us simplify the integrand. This is where the magic happens.

Expanding the square, we get $(x - \frac{1}{2})^2 = x^2 - x + \frac{1}{4}$. When we subtract this from $(1 - x)$, the linear terms $-x$ and $+x$ cancel out perfectly.

We are left with the elegant expression:

$$\frac{3}{4} - x^2$$

This is the soul of the problem. What looked like a complex subtraction of two functions has collapsed into a simple, beautiful quadratic. We are now integrating:

$$\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \left( \frac{3}{4} - x^2 \right) dx$$

### Phase 4: The Final Integration

Applying the power rule, the integral becomes:

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

Now, we substitute the limits with precision. For the upper limit $x = \frac{\sqrt{3}}{2}$, we calculate:

$$\left( \frac{3}{4} \cdot \frac{\sqrt{3}}{2} \right) - \frac{1}{3}\left( \frac{\sqrt{3}}{2} \right)^3 = \frac{3\sqrt{3}}{8} - \frac{3\sqrt{3}}{24} = \frac{3\sqrt{3}}{8} - \frac{\sqrt{3}}{8} = \frac{\sqrt{3}}{4}$$

For the lower limit $x = \frac{1}{2}$, we calculate:

$$\left( \frac{3}{4} \cdot \frac{1}{2} \right) - \frac{1}{3}\left( \frac{1}{2} \right)^3 = \frac{3}{8} - \frac{1}{24} = \frac{1}{3}$$

Subtracting the lower from the upper, we arrive at our destination:

**$$A = \frac{\sqrt{3}}{4} - \frac{1}{3}$$**

### Reflection

Look at what you have achieved. You navigated the piecewise definition, identified the correct interval, simplified a daunting integrand through algebraic cancellation, and executed the definite integral with care.

This is the essence of JEE Advanced physics and math—not brute force, but the art of simplifying the complex until only the truth remains. Keep this clarity, keep this focus, and you will conquer any problem that comes your way.

Select Answer:

Visualized Solution

Visualizing the Functions

Defining $f (x)$ in the Interval

Analyzing $g (x)$ and Intersection

Determining the Upper Curve

Simplifying the Integrand

Setting up the Definite Integral

Substituting the Upper Limit

Substituting the Lower Limit

Final Calculation

Key Takeaway

The Sigma Insight: Area Bounded by Curves

The Geometry of Hidden Elegance

Phase 1

Phase 2

Phase 3

Phase 4

Reflection

Similar Questions

Let the functions $f : R \to R$ and $g : R \to R$ be defined by $f (x) = e^{x - 1} - e^{- ∣ x - 1∣}$ and $g (x) = \frac{1}{2} (e^{x - 1} + e^{1 - x})$ . Then the area of the region in the first quadrant bounded by the curves $y = f (x)$ , $y = g (x)$ and $x = 0$ is

The area bounded by the curves $y = ∣ x^{2} - 1∣$ and $y = 1$ is

Find the area of the region bounded by the curves $y = x^{2}, y = 2 - x^{2}$ and $y = 2$ , which lies to the right of the line $x = 1$ .

Sketch the region bounded by the curves $y = x^{2}$ and $y = \frac{2}{1 + x ^{2}}$ . Find the area.

The area (in sq. units) of the region ${(x, y) : x \geq 0, x + y \leq 3, x^{2} \leq 4 y and y \leq 1 + x}$ is:

The area of the region between the curves $y = \frac{1 + s i n x}{c o s x}$ and $y = \frac{1 - s i n x}{c o s x}$ bounded by the lines $x = 0$ and $x = π /4$ is

Find the area bounded by the curves $x^{2} = y, x^{2} = - y$ and $y^{2} = 4 x - 3$ .

The area of the region bounded by the curves $x (1 + y^{2}) = 1$ and $y^{2} = 2 x$ is:

Let $f (x) =$ Maximum ${x^{2}, (1 - x)^{2}, 2 x (1 - x)}$ , where $0 \leq x \leq 1$ . Determine the area of the region bounded by the curves $y = f (x), x$ -axis, $x = 0$ and $x = 1$ .

The area of the region ${(x, y) : 0 \leq x \leq \frac{9}{4}, 0 \leq y \leq 1, x \leq 3 y, x + y \geq 2}$ is