Let (a, b)$(a, b)$ be the point of intersection of the curve x^2 = 2y$x^2 = 2y$ and the straight line y - 2x - 6 = 0$y - 2x - 6 = 0$ in the second quadrant. Then the integral I = int_a^b frac9x^21 + 5^x \, dx$I = \int_{a}^{b} \frac{9x^2}{1 + 5^x} \, dx$ is equal to:

### Related Formula textKing's Property: int_a^b f(x) dx = int_a^b f(a+b-x) dx$$\text{King's Property: } \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$$ ### Core Logic First, we find the coordinates of intersection in the second quadrant to determine the integration limits a$a$ and b$b$. ### Step 1: Find points of intersection Substitute y = fracx^22$y = \frac{x^2}{2}$ into the line equation y - 2x - 6 = 0$y - 2x - 6 = 0$: fracx^22 - 2x - 6 = 0 implies x^2 - 4x - 12 = 0$$\frac{x^2}{2} - 2x - 6 = 0 \implies x^2 - 4x - 12 = 0$$ (x - 6)(x + 2) = 0 implies x = 6 quad textor quad x = -2$$(x - 6)(x + 2) = 0 \implies x = 6 \quad \text{or} \quad x = -2$$ Since the point (a, b)$(a, b)$ lies in the second quadrant, x$x$ must be negative: a = -2$a = -2$ b = 2(a) + 6 = 2(-2) + 6 = 2$$b = 2(a) + 6 = 2(-2) + 6 = 2$$ Thus, the integration limits are a = -2$a = -2$ and b = 2$b = 2$. ### Step 2: Solve the Integral using King's Property The integral is: I = int_-2^2 frac9x^21 + 5^x dx quad text--- (1)$$I = \int_{-2}^{2} \frac{9x^2}{1 + 5^x} dx \quad \text{--- (1)}$$ Apply King's property, substituting x to -x$x \to -x$ (since -2 + 2 - x = -x$-2 + 2 - x = -x$): I = int_-2^2 frac9(-x)^21 + 5^-x dx = int_-2^2 frac9x^2 cdot 5^x1 + 5^x dx quad text--- (2)$$I = \int_{-2}^{2} \frac{9(-x)^2}{1 + 5^{-x}} dx = \int_{-2}^{2} \frac{9x^2 \cdot 5^x}{1 + 5^x} dx \quad \text{--- (2)}$$ Adding equations (1) and (2): 2I = int_-2^2 9x^2 left( frac1 + 5^x1 + 5^x right) dx = int_-2^2 9x^2 dx$$2I = \int_{-2}^{2} 9x^2 \left( \frac{1 + 5^x}{1 + 5^x} \right) dx = \int_{-2}^{2} 9x^2 dx$$ Since 9x^2$9x^2$ is an even function: 2I = 2 int_0^2 9x^2 dx implies I = left[ 3x^3 right]_0^2 = 3(8) - 0 = 24$$2I = 2 \int_{0}^{2} 9x^2 dx \implies I = \left[ 3x^3 \right]_{0}^{2} = 3(8) - 0 = 24$$ ### Pattern Recognition Whenever you see an exponential denominator like 1 + c^x$1 + c^x$ inside a symmetric interval definite integral [-a, a]$[-a, a]$, applying King's property will almost always cancel the exponential factor cleanly when the remaining numerator is even. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integral Calculus