The area (in sq. units) of the region {x ∈ R : x ≥ 0, y ≥ 0, y ≥ x–2 and y ≤ √x }, is
(1) 10/3
(2) 1

Q: Why did we reject the intersection point at $x = 1$?

Great catch! When we squared the equation $\sqrt{x} = x - 2$, we introduced an extraneous root. If you plug $x = 1$ back into the original equation, you get $\sqrt{1} = 1 - 2$, which is $1 = -1$. This is clearly false! Always verify your intersection points in the original equations.

Q: Why is integrating with respect to $y$ better than $x$?

It is all about elegance and efficiency. If you integrate with respect to $x$, you must calculate $\int_{0}^{2} \sqrt{x} dx + \int_{2}^{4} (\sqrt{x} - (x - 2)) dx$. By integrating with respect to $y$, you treat the region as one single block, avoiding the risk of arithmetic errors in splitting the integral.

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

Think of the region as a horizontal slice. As you move from left to right at any height $y$, you hit the parabola $x = y^2$ first, and then the line $x = y + 2$. Therefore, the line is the right boundary and the parabola is the left boundary.

Q: What if I forgot the $y \geq 0$ constraint?

If you ignore the constraints, you might include regions outside the first quadrant, leading to an incorrect area. The constraints define the 'fencing' of your problem; always check them before you start integrating!

Q: Is the area always positive?

Yes! Area is a geometric magnitude. If you ever calculate an integral and get a negative result, it means you likely swapped the 'right' and 'left' functions. Just take the absolute value, but better yet, ensure your integrand is always (Right - Left).

Question

The area (in sq. units) of the region $\{x \in R : x \geq 0, y \geq 0, y \geq x - 2 	ext{ and } y \leq \sqrt{x} \}$, is

Options:

(A) 10/3

(B) 13/3

(C) 5/3

(D) 8/3

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane restricted to the first quadrant, where $x \geq 0$ and $y \geq 0$. Within this space, we have two distinct mathematical entities: a graceful, curving parabola defined by $y = \sqrt{x}$ and a sharp, linear path defined by $y = x - 2$.

Our goal is to find the area of the region trapped between them. This is a study of the geometric dance between these two functions.

### The Intersection: Finding the Meeting Point

Before we can calculate the area, we must determine where our boundaries begin and end. We set the curves equal to each other: $\sqrt{x} = x - 2$.

Squaring both sides gives us $x = (x - 2)^2$, which expands to $x = x^2 - 4x + 4$. Rearranging this into a quadratic equation, we obtain:

$$x^2 - 5x + 4 = 0$$

Factoring this, we find $(x - 4)(x - 1) = 0$.

We have two potential solutions: $x = 1$ and $x = 4$. However, we must check these against the original equation $\sqrt{x} = x - 2$.

If we plug in $x = 1$, we get $1 = -1$, which is impossible. This is a classic trap in coordinate geometry—squaring an equation can create "ghost" solutions.

We discard $x = 1$ and keep $x = 4$. At $x = 4$, we find $y = 2$. This is our primary intersection point.

### The Strategy: Horizontal vs. Vertical

We must now decide how to measure this area. If we look at the region vertically, the "bottom" boundary changes.

From $x = 0$ to $x = 2$, the bottom is the x-axis ($y = 0$). From $x = 2$ to $x = 4$, the bottom is the line $y = x - 2$. This would force us to split our work into two separate integrals.

There is a more elegant path. If we look at the region horizontally, we see a single, continuous strip.

By integrating with respect to $y$, we define the right boundary as $x = y + 2$ and the left boundary as $x = y^2$. This allows us to express the entire area as a single, beautiful integral:

$$Area = \int_{0}^{2} ((y + 2) - y^2) \, dy$$

### The Integration: Bringing it Home

Now, we perform the calculus by integrating term by term:

$$\int_{0}^{2} (y + 2 - y^2) \, dy = \left[ \frac{y^2}{2} + 2y - \frac{y^3}{3} \right]_{0}^{2}$$

We substitute our limits into the expression. Plugging in $y = 2$, we get:

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

Simplifying this, we have $6 - \frac{8}{3}$. Finding a common denominator, we get:

$$\frac{18 - 8}{3} = \frac{10}{3}$$

### The Conclusion

The area of our region is exactly **$\frac{10}{3}$ square units**.

By choosing the right perspective—integrating with respect to $y$—we turned a potentially messy two-part problem into a single, elegant calculation. In JEE Advanced, the most powerful tool you have is the ability to choose the path that makes the math reveal its own simplicity.

Animated Solution for Mathematics - Definite Integration: The area (in sq. units) of the region ${x \in R : x \geq 0, y \geq 0, y \geq x - 2 and y \leq x}$ , is

Select Answer:

Visualized Solution

Analyze the Constraints

Plotting $y = x$

Plotting the Line $y = x - 2$

Finding the Intersection

Identifying the Bounded Region

Choosing the Integration Strategy

Setting up the Integral

Performing the Integration

Applying the Limits

Final Calculation

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Intersection

The Strategy

The Integration

The Conclusion

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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∈R:x≥0,y≥0,y≥x−2 and y≤x​}, is