### 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

### Related Formula int sqrta^2 + x^2 dx = fracx2 sqrta^2 + x^2 + fraca^22 lnleft| x + sqrta^2 + x^2 right|$\int \sqrt{a^2 + x^2} dx = \frac{x}{2} \sqrt{a^2 + x^2} + \frac{a^2}{2} \ln\left| x + \sqrt{a^2 + x^2} \right|$ ### Core Logic We first rationalise the denominator to split the integral into two standard integration terms. ### Step 1: Rationalise the integrand Multiply the numerator and denominator by sqrt3+x^2 - sqrt1+x^2$\sqrt{3+x^2} - \sqrt{1+x^2}$: frac1sqrt3 + x^2 + sqrt1 + x^2 = fracsqrt3 + x^2 - sqrt1 + x^2(3+x^2) - (1+x^2) = fracsqrt3 + x^2 - sqrt1 + x^22$\frac{1}{\sqrt{3 + x^2} + \sqrt{1 + x^2}} = \frac{\sqrt{3 + x^2} - \sqrt{1 + x^2}}{(3+x^2) - (1+x^2)} = \frac{\sqrt{3 + x^2} - \sqrt{1 + x^2}}{2}$ Thus, the integral expression simplifies to: I = 4 int_0^1 left( fracsqrt3 + x^2 - sqrt1 + x^22 right) dx - 3log_eleft(sqrt3right)$I = 4 \int_{0}^{1} \left( \frac{\sqrt{3 + x^2} - \sqrt{1 + x^2}}{2} \right) dx - 3\log_{e}\left(\sqrt{3}\right)$ I = 2 int_0^1 sqrt3 + x^2 dx - 2 int_0^1 sqrt1 + x^2 dx - frac32 log_e 3$I = 2 \int_{0}^{1} \sqrt{3 + x^2} dx - 2 \int_{0}^{1} \sqrt{1 + x^2} dx - \frac{3}{2} \log_{e} 3$ ### Step 2: Evaluate the integrals For the first integral: 2 int_0^1 sqrt3 + x^2 dx = 2 left[ fracx2 sqrt3 + x^2 + frac32 lnleft| x + sqrt3 + x^2 right| right]_0^1$2 \int_{0}^{1} \sqrt{3 + x^2} dx = 2 \left[ \frac{x}{2} \sqrt{3 + x^2} + \frac{3}{2} \ln\left| x + \sqrt{3 + x^2} \right| \right]_{0}^{1}$ = left[ x sqrt3 + x^2 + 3 lnleft| x + sqrt3 + x^2 right| right]_0^1$= \left[ x \sqrt{3 + x^2} + 3 \ln\left| x + \sqrt{3 + x^2} \right| \right]_{0}^{1}$ = left( sqrt4 + 3 ln(1 + sqrt4) right) - left( 0 + 3 lnsqrt3 right)$= \left( \sqrt{4} + 3 \ln(1 + \sqrt{4}) \right) - \left( 0 + 3 \ln\sqrt{3} \right)$ = 2 + 3 ln 3 - frac32 ln 3 = 2 + frac32 ln 3$= 2 + 3 \ln 3 - \frac{3}{2} \ln 3 = 2 + \frac{3}{2} \ln 3$ For the second integral: -2 int_0^1 sqrt1 + x^2 dx = -2 left[ fracx2 sqrt1 + x^2 + frac12 lnleft| x + sqrt1 + x^2 right| right]_0^1$-2 \int_{0}^{1} \sqrt{1 + x^2} dx = -2 \left[ \frac{x}{2} \sqrt{1 + x^2} + \frac{1}{2} \ln\left| x + \sqrt{1 + x^2} \right| \right]_{0}^{1}$ = - left[ x sqrt1 + x^2 + lnleft| x + sqrt1 + x^2 right| right]_0^1$= - \left[ x \sqrt{1 + x^2} + \ln\left| x + \sqrt{1 + x^2} \right| \right]_{0}^{1}$ = - left( sqrt2 + ln(1 + sqrt2) right)$= - \left( \sqrt{2} + \ln(1 + \sqrt{2}) \right)$ ### Step 3: Sum the terms Now compile all terms: I = left( 2 + frac32 ln 3 right) - sqrt2 - ln(1 + sqrt2) - frac32 ln 3$I = \left( 2 + \frac{3}{2} \ln 3 \right) - \sqrt{2} - \ln(1 + \sqrt{2}) - \frac{3}{2} \ln 3$ I = 2 - sqrt2 - ln(1 + sqrt2)$I = 2 - \sqrt{2} - \ln(1 + \sqrt{2})$ ### Pattern Recognition Integration of roots of quadratics: Always look for algebraic rationalisation when dealing with sum of root denominators. It directly reduces complex fractions into standard integrable functions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integral Calculus

### Related Formula Area enclosed between two symmetric functions across four quadrants: alpha = 4 times left( textArea of circle in 1^textst text quadrant - int_0^1 y_textparabola \, dx right)$\alpha = 4 \times \left( \text{Area of circle in } 1^{\text{st}} \text{ quadrant} - \int_{0}^{1} y_{\text{parabola}} \, dx \right)$ ### Core Logic The curves are symmetric across the axes. The region is enclosed between the unit circle x^2 + y^2 = 1$x^2 + y^2 = 1$ and the parabolas y = pm(1 - x^2)$y = \pm(1 - x^2)$.

By symmetry, the area alpha$\alpha$ in the first quadrant is the area under the circle minus the area under the parabola from x=0$x=0$ to x=1$x=1$. ### Step 1: Calculate the Area Components Area of a circle quadrant with r=1$r=1$ is fracpi4$\frac{\pi}{4}$. Area under the parabola: int_0^1 (1 - x^2) \, dx = left[ x - fracx^33 right]_0^1 = 1 - frac13 = frac23$\int_{0}^{1} (1 - x^2) \, dx = \left[ x - \frac{x^3}{3} \right]_{0}^{1} = 1 - \frac{1}{3} = \frac{2}{3}$ ### Step 2: Evaluate the Total Area alpha = 4 left[ fracpi4 - frac23 right] = pi - frac83$\alpha = 4 \left[ \frac{\pi}{4} - \frac{2}{3} \right] = \pi - \frac{8}{3}$ Multiplying both sides by 9: 9alpha = 9pi - 24$9\alpha = 9\pi - 24$ Comparing with 9alpha = betapi + gamma$9\alpha = \beta\pi + \gamma$, we get: beta = 9, quad gamma = -24$\beta = 9, \quad \gamma = -24$ Hence, |beta - gamma| = |9 - (-24)| = 33$|\beta - \gamma| = |9 - (-24)| = 33$ ### Pattern Recognition Symmetry helps divide the integration work by 4. Recognizing standard shapes (like a circle sector) allows you to bypass complex integration entirely for half of the problem. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integral Calculus

### Related Formula The area enclosed between upper bounding function y_textupper$y_{\text{upper}}$ and lower function y_textlower$y_{\text{lower}}$ is: textArea = int_a^b (y_textupper - y_textlower) \, dx$\text{Area} = \int_{a}^{b} (y_{\text{upper}} - y_{\text{lower}}) \, dx$ ### Core Logic We need to find the intersection points of the curves to understand the min\x+7, 11-3x\$\min\{x+7, 11-3x\}$ boundary transition: 1) x+7 = 11-3x implies 4x = 4 implies x = 1$x+7 = 11-3x \implies 4x = 4 \implies x = 1$. Hence, the line switches behavior at x=1$x=1$. 2) Intersecting 1+x^2$1+x^2$ with x+7$x+7$: x^2 - x - 6 = 0 implies (x-3)(x+2) = 0 implies x = -2 quad textor quad x = 3$x^2 - x - 6 = 0 \implies (x-3)(x+2) = 0 \implies x = -2 \quad \text{or} \quad x = 3$ 3) Intersecting 1+x^2$1+x^2$ with 11-3x$11-3x$: x^2 + 3x - 10 = 0 implies (x+5)(x-2) = 0 implies x = 2 quad textor quad x = -5$x^2 + 3x - 10 = 0 \implies (x+5)(x-2) = 0 \implies x = 2 \quad \text{or} \quad x = -5$

### Step 1: Set up Integrals The transition points show that from x = -2$x = -2$ to 1$1$, the upper line is x+7$x+7$, and from x = 1$x = 1$ to 2$2$, the upper line is 11-3x$11-3x$. A = int_-2^1 ((x + 7) - (1 + x^2)) \, dx + int_1^2 ((11 - 3x) - (1 + x^2)) \, dx$A = \int_{-2}^{1} ((x + 7) - (1 + x^2)) \, dx + \int_{1}^{2} ((11 - 3x) - (1 + x^2)) \, dx$ A = int_-2^1 (x + 6 - x^2) \, dx + int_1^2 (10 - 3x - x^2) \, dx$A = \int_{-2}^{1} (x + 6 - x^2) \, dx + \int_{1}^{2} (10 - 3x - x^2) \, dx$ ### Step 2: Integration Evaluation Evaluating the first integral: left[ fracx^22 + 6x - fracx^33 right]_-2^1 = left(frac12 + 6 - frac13right) - left(2 - 12 + frac83right) = frac376 - left(-frac223right) = frac272$\left[ \frac{x^2}{2} + 6x - \frac{x^3}{3} \right]_{-2}^{1} = \left(\frac{1}{2} + 6 - \frac{1}{3}\right) - \left(2 - 12 + \frac{8}{3}\right) = \frac{37}{6} - \left(-\frac{22}{3}\right) = \frac{27}{2}$ Evaluating the second integral: left[ 10x - frac3x^22 - fracx^33 right]_1^2 = left(20 - 6 - frac83right) - left(10 - frac32 - frac13right) = frac343 - frac496 = frac196$\left[ 10x - \frac{3x^2}{2} - \frac{x^3}{3} \right]_{1}^{2} = \left(20 - 6 - \frac{8}{3}\right) - \left(10 - \frac{3}{2} - \frac{1}{3}\right) = \frac{34}{3} - \frac{49}{6} = \frac{19}{6}$ Total Area A$A$: A = frac272 + frac196 = frac81 + 196 = frac1006 = frac503$A = \frac{27}{2} + \frac{19}{6} = \frac{81 + 19}{6} = \frac{100}{6} = \frac{50}{3}$ ### Step 3: Calculate 3A Multiplying the total area by 3: 3A = 3 cdot left(frac503right) = 50$3A = 3 \cdot \left(\frac{50}{3}\right) = 50$ ### Pattern Recognition When a boundary contains a min\\$\min\{\}$ or max\\$\max\{\}$ component, always solve for their internal intersection first to identify the exact splitting point of your definite integrals. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integral Calculus