The area bounded by the curves y = √x, 2y + 3 = x and x-axis in the 1st quadrant is (a) 9 (b) 27/4 (

Q: Why can't we just use vertical strips ($dx$) to find the area?

You absolutely can, but it is a trap! If you use vertical strips, you must split the integral into two parts: one from $x = 0$ to $x = 3$ (where the area is bounded by the parabola and the x-axis) and another from $x = 3$ to $x = 9$ (where the area is bounded by the line and the x-axis). Integrating with respect to $y$ allows you to solve it in one single, clean step, minimizing the chance of arithmetic errors.

Q: Why do we reject $y = -1$ when solving the quadratic equation?

Great catch! The problem explicitly restricts us to the 1st quadrant. In the 1st quadrant, both $x$ and $y$ must be non-negative. Since $y = -1$ falls outside this domain, it is physically irrelevant to our area calculation. Always check your constraints before finalizing your limits of integration.

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

Think of the region as a slice of bread. If you draw a horizontal line through the region, the point where it hits the curve first (the smaller $x$ value) is your left boundary, and the point where it hits the curve last (the larger $x$ value) is your right boundary. In our case, for any given $y$, the line $x = 2y + 3$ is further to the right than the parabola $x = y^2$.

Question

The area bounded by the curves $y = \sqrt{x}$, $2y + 3 = x$ and x-axis in the 1st quadrant is

Options:

(A) 9

(B) 27/4

(C) 36

(D) 18

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When you look at a problem involving curves like $y = \sqrt{x}$ and $2y + 3 = x$, I want you to stop and visualize. Do not rush to the algebra.

See the parabola opening its arms along the $x$-axis, and see the line cutting through the plane. Our goal is to find the area trapped between them in the 1st quadrant.

### The Algebraic Intersection

Before we can integrate, we must define our limits. Where do these two paths meet?

We set $x = y^2$ and $x = 2y + 3$. By equating them, we arrive at the quadratic equation:

$$y^2 - 2y - 3 = 0$$

Factoring this, we find $(y - 3)(y + 1) = 0$. Here is where the discipline of a JEE aspirant shines.

We have two roots, $y = 3$ and $y = -1$. But we are in the 1st quadrant! The negative root is a mathematical ghost—it exists, but it does not belong in our physical region.

We discard it and keep $y = 3$. This tells us our upper limit of integration is $y = 3$.

### The Strategic Pivot

Now, the most critical decision: vertical or horizontal strips? If you choose vertical strips ($dx$), you are walking into a labyrinth.

You would have to split the integral at $x = 3$ because the lower boundary changes from the parabola to the $x$-axis. Why make life harder?

If we use horizontal strips ($dy$), the strip always touches the parabola on the left and the line on the right, from $y = 0$ all the way to $y = 3$. This is the beauty of choosing the right perspective.

We set up our integral as $A = \int_{0}^{3} (x_{right} - x_{left}) \, dy$. Substituting our functions, we get:

$$A = \int_{0}^{3} ((2y + 3) - y^2) \, dy$$

### The Execution

Now, we integrate. This is the moment of truth.

We apply the power rule term by term: the integral of $2y$ is $y^2$, the integral of $3$ is $3y$, and the integral of $y^2$ is $\frac{y^3}{3}$. We evaluate this from $0$ to $3$:

$$A = \left[ y^2 + 3y - \frac{y^3}{3} \right]_0^3$$

Plugging in the upper limit, we get $(3^2 + 3(3) - \frac{3^3}{3})$, which simplifies to $9 + 9 - 9$.

The result is **$9$**. The lower limit of $0$ contributes nothing.

And there it is—**$9$ square units**. It is clean, it is precise, and it is elegant. Remember, in JEE Advanced, the smartest path is often the one that simplifies the problem before you even pick up your pen.

Animated Solution for Mathematics - Definite Integration: The area bounded by the curves $y = x$ , $2 y + 3 = x$ and x-axis in the 1st quadrant is

Select Answer:

Visualized Solution

Identify the Boundaries

Intersection of Curve and Line

Solve for $y$

Valid Intersection Point

Line Intercept on x-axis

The Bounded Region

Vertical vs Horizontal Strips

Set up the Integral

Substitute the Boundaries

Integrate the Expression

Evaluate the Limits

Final Answer

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Algebraic Intersection

The Strategic Pivot

The Execution

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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 bounded by the curves y=x​, 2y+3=x and x-axis in the 1st quadrant is