Let (a, b) be the point of intersection of the curve x² = 2y and the straight line y - 2x - 6 = 0 in

Q: Why do we reject $x=6$ when solving the quadratic?

The problem explicitly states the intersection is in the second quadrant. In the Cartesian plane, the second quadrant is defined by $x 0$. Since $6 > 0$, it lies in the first quadrant. Always respect the constraints given in the problem statement!

Q: Can we use this property for any integral?

This property, often called the 'King's Property' or the 'Symmetric Limit Property', is specifically powerful when the denominator is of the form $a^x + 1$ or similar. It doesn't work for every integral, but it is your best friend when you see symmetric limits and a 'messy' denominator that looks impossible to integrate directly.

Q: Why can't we just integrate $\frac{9x^2}{1+5^x}$ directly?

You could try, but integrating $\frac{9x^2}{1+5^x}$ directly is extremely difficult because there is no elementary anti-derivative for that form. The property is designed to bypass the need for direct integration by simplifying the integrand itself into a simple polynomial.

Question

Let $(a, b)$ be the point of intersection of the curve $x^2 = 2y$ and the straight line $y - 2x - 6 = 0$ in the second quadrant. Then the integral $I = \int_a^b rac{9x^2}{1+5^x} dx$ is equal to :

Options:

(A) 24

(B) 27

(C) 18

(D) 21

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To begin, we visualize the geometry of the given parabola, $x^2 = 2y$, and the line, $y - 2x - 6 = 0$. Our first mission is to find their points of intersection.

Substituting $y = 2x + 6$ into the parabola equation, we obtain:

$$x^2 = 2(2x + 6) \Rightarrow x^2 - 4x - 12 = 0$$

Factoring this quadratic equation, we find:

$$(x - 6)(x + 2) = 0$$

This yields two potential $x$-coordinates: $x = 6$ and $x = -2$. Since the problem specifies the intersection in the second quadrant, where $x$ must be negative, we reject $x = 6$ and accept $x = -2$.

Plugging $x = -2$ back into the line equation, we find $y = 2(-2) + 6 = 2$. Thus, our intersection point is $(-2, 2)$, providing the limits of integration as $a = -2$ and $b = 2$.

### The Integral: A Symmetric Challenge

We now face the integral:

$$I = \int_{-2}^{2} \frac{9x^2}{1+5^x} dx$$

Whenever you encounter an integral with symmetric limits $\int_{-a}^{a} f(x) dx$, your intuition should immediately jump to the property:

$$\int_{-a}^{a} f(x) dx = \int_{0}^{a} [f(x) + f(-x)] dx$$

This is a powerful tool in your JEE arsenal. It allows us to transform a complex integrand into something manageable. Let us define our function as $f(x) = \frac{9x^2}{1+5^x}$.

### The Magic of the King's Property

To apply the property, we calculate $f(-x)$:

$$f(-x) = \frac{9(-x)^2}{1+5^{-x}} = \frac{9x^2}{1 + \frac{1}{5^x}}$$

Multiplying the numerator and denominator by $5^x$, we simplify this to:

$$f(-x) = \frac{9x^2 \cdot 5^x}{5^x + 1}$$

Now, we add $f(x)$ and $f(-x)$:

$$f(x) + f(-x) = \frac{9x^2}{1+5^x} + \frac{9x^2 \cdot 5^x}{1+5^x} = \frac{9x^2(1+5^x)}{1+5^x}$$

The term $(1+5^x)$ cancels out beautifully, leaving us with just $9x^2$. This is the "Aha!" moment where the complex exponential term vanishes.

### The Final Victory

Our integral has collapsed from a daunting expression into a simple polynomial:

$$I = \int_{0}^{2} 9x^2 dx$$

The anti-derivative of $9x^2$ is $3x^3$. Evaluating this from $0$ to $2$:

$$I = [3x^3]_{0}^{2} = 3(2)^3 - 3(0)^3 = 3(8) = 24$$

Through the power of geometric visualization and the elegance of definite integral properties, we have tamed the beast. The final answer is **24**.

Animated Solution for Mathematics - Definite Integration: Let $(a, b)$ be the point of intersection of the curve $x^{2} = 2 y$ and the straight line $y - 2 x - 6 = 0$ in the second quadrant. Then the integral $I = \int_a^b rac{9x^2}{1+5^x} dx$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Curves

Finding Intersection Points

Setting up the Equation

Solving the Quadratic

Identifying Point $(a, b)$

Setting up the Integral

King's Rule for Symmetric Limits

Evaluating $f (- x)$

Simplifying $f (- x)$

Adding $f (x)$ and $f (- x)$

Integrating the Simplified Function

Final Calculation

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Integral

The Magic of the King's Property

The Final Victory

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Select Answer:

Visualized Solution

Visualizing the Curves

Finding Intersection Points

Setting up the Equation

Solving the Quadratic

Identifying Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (a,b)

Setting up the Integral

King's Rule for Symmetric Limits

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(−x)

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(−x)

Adding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(−x)