Let the curve y = y(x) be the solution of the differential equation, dy/dx = 2(x + 1). If the numeri

Q: Wait, why do we need the roots to be real? Can't the area be imaginary?

Great question! The problem specifies a 'numerical value of area bounded by the curve'. In geometry, area must be a real, positive quantity. If the roots were complex, the parabola would never touch the x-axis, and no bounded area would exist. Thus, we must ensure the discriminant $D > 0$.

Q: Why can't I just integrate the function from $-\infty$ to $+\infty$?

Because that integral would diverge! The area is only bounded between the two x-intercepts where the curve dips below the x-axis. We are looking for the finite region trapped between the curve and the axis.

Q: Is the shortcut formula $Area = \frac{|a|}{6}(\beta - \alpha)^3$ always valid?

Yes, for any parabola $y = ax^2 + bx + c$ intersecting the x-axis at $\alpha$ and $\beta$, the area of the region between the curve and the x-axis is exactly $\frac{|a|}{6}(\beta - \alpha)^3$. It is a powerful tool in your JEE arsenal to save precious time!

Question

Let the curve $y = y(x)$ be the solution of the differential equation, $\frac{dy}{dx} = 2(x + 1)$. If the numerical value of area bounded by the curve $y = y(x)$ and $x$-axis is $\frac{4\sqrt{8}}{3}$, then the value of $y(1)$ is equal to ____

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When you see the differential equation $\frac{dy}{dx} = 2(x + 1)$, do not just see symbols. See a rate of change. This equation describes how the slope of the curve evolves as we move along the $x$-axis.

To find the curve, we integrate the expression. Separating the variables gives us:

$$\int dy = \int 2(x + 1) dx$$

This leads us to the general solution:

$$y = x^2 + 2x + C$$

This represents a family of parabolas, all identical in shape but shifting vertically based on the constant $C$.

### The Mystery of the Bounded Area

For a parabola to bound an area with the $x$-axis, it must cross the axis at two distinct points. We find these roots by solving $x^2 + 2x + C = 0$.

Using the quadratic formula, we find the roots $\alpha$ and $\beta$:

$$\alpha, \beta = -1 \pm \sqrt{1 - C}$$

These roots serve as our boundaries for the area calculation.

### The Power of the Shortcut

While you could set up the definite integral $\int_{\alpha}^{\beta} (x^2 + 2x + C) dx$, time is your most precious resource in a JEE Advanced exam. Instead, we use the geometric property of parabolas: the area between a parabola and the $x$-axis is given by:

$$Area = \frac{|a|}{6}(\beta - \alpha)^3$$

Here, $a = 1$. The difference between the roots $\beta - \alpha$ is simply $2\sqrt{1 - C}$.

Plugging this into our formula, we obtain:

$$Area = \frac{1}{6}(2\sqrt{1 - C})^3 = \frac{4}{3}(1 - C)^{3/2}$$

### The Final Reveal

We equate this expression to the given area, $\frac{4\sqrt{8}}{3}$. The $\frac{4}{3}$ cancels out, leaving us with:

$$(1 - C)^{3/2} = \sqrt{8}$$

Since $\sqrt{8} = 8^{1/2} = (2^3)^{1/2} = 2^{3/2}$, we immediately see that $1 - C = 2$. This implies that **$C = -1$**.

Our curve is now fully defined as $y = x^2 + 2x - 1$. Finally, finding $y(1)$ is a matter of simple substitution:

$$y(1) = (1)^2 + 2(1) - 1 = 2$$

**The final answer is 2.**

Enter Numerical Value:

Visualized Solution

Analyze the Differential Equation

Separate Variables and Integrate

Find the General Solution $y (x)$

Identify the Geometric Shape

Find the $x$ -intercepts

Define the Area Integral

Apply the Parabolic Area Formula

Equate to the Given Area

Solve for the Constant $C$

Final Equation of the Curve

Calculate $y (1)$

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Mystery of the Bounded Area

The Power of the Shortcut

The Final Reveal

Similar Questions

If $y (x)$ is the solution of the differential equation $x d y - (y^{2} - 4 y) d x = 0$ for $x > 0$ , $y (1) = 2$ , and the slope of the curve $y = y (x)$ is never zero, then the value of $10 y (2)$ is ______.

If the curve $y = y (x)$ represented by the solution of the differential equation $(2 x y^{2} - y) d x + x d x = 0$ , passes through the intersection of the lines, $2 x - 3 y = 1$ and $3 x + 2 y = 8$ , then $∣ y (1) ∣$ is equal to

If $y = y (x)$ is the solution of the differential equation, $e^{y} (\frac{d y}{d x} - 1) = e^{x}$ such that $y (0) = 0$ , then $y (1)$ is equal to

Let $y = y (x)$ be the solution of the differential equation $e^{x} 1 - y^{2} d x + (\frac{y}{x}) d y = 0, y (1) = - 1$ . Then the value of $(y (3))^{2}$ is equal to:

Let $\int_{0}^{x} 1 - (y^{'} (t))^{2} d t = \int_{0}^{x} y (t) d t, 0 \leq x \leq 3, y \geq 0, y (0) = 0$ . Then at $x = 2, y^{''} + y + 1$ is equal to

If $(2 + sin x) \frac{d y}{d x} + (y + 1) cos x = 0$ and $y (0) = 1$ , then $y (\frac{π}{2})$ is equal to:

If $\frac{d y}{d x} = \frac{2 ^{x} y + 2 ^{y} \cdot 2 ^{x}}{2 ^{x} + 2 ^{x + y} l o g _{e} 2}, y (0) = 0$ , then for $y = 1$ , the value of $x$ lies in the interval:

If the solution $y = y (x)$ of the differential equation $(x^{4} + 2 x^{3} + 3 x^{2} + 2 x + 2) d y - (2 x^{2} + 2 x + 3) d x = 0$ satisfies $y (- 1) = - \frac{π}{4}$ , then $y (0)$ is equal to :

If $\frac{d y}{d x} + \frac{2 ^{x - y} ( 2 ^{y} - 1 )}{2 ^{x} - 1} = 0, x, y > 0, y (1) = 1$ , then $y (2)$ is equal to :