4int_0^1left(frac1sqrt3 + x^2 + sqrt1 + x^2right)mathrmdx - 3log_eleft(sqrt3right)$4\int_{0}^{1}\left(\frac{1}{\sqrt{3 + x^2} + \sqrt{1 + x^2}}\right)\mathrm{d}x - 3\log_{e}\left(\sqrt{3}\right)$ 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

### Related Formula Greatest Integer Function boundaries: [f(x)] = k quad textfor quad k le f(x) < k+1, \ k in mathbbZ$$[f(x)] = k \quad \text{for} \quad k \le f(x) < k+1, \ k \in \mathbb{Z}$$ ### Core Logic Analyze the value variations of f(x) = e^1-x$f(x) = e^{1-x}$ across the integration limits [0, e^3]$[0, e^3]$ to break down the integral into distinct piecewise continuous intervals. ### Step 1: Determine Step Function Transition Points Let y = e^1-x$y = e^{1-x}$. * At x = 0 implies y = e^1 approx 2.718$x = 0 \implies y = e^1 \approx 2.718$ * As x$x$ increases, e^1-x$e^{1-x}$ decreases monotonically. * Find x$x$ where y = 2 implies e^1-x = 2 implies 1-x = ln 2 implies x = 1 - ln 2$y = 2 \implies e^{1-x} = 2 \implies 1-x = \ln 2 \implies x = 1 - \ln 2$. * Find x$x$ where y = 1 implies e^1-x = 1 implies 1-x = 0 implies x = 1$y = 1 \implies e^{1-x} = 1 \implies 1-x = 0 \implies x = 1$. * At the final boundary x = e^3 implies y = e^1-e^3$x = e^3 \implies y = e^{1-e^3}$, which is a very small positive decimal strictly inside (0,1)$(0,1)$. ### Step 2: Split the Definite Integral Rewrite the integral based on the isolated interval blocks: I = int_0^1-ln 2 2 \, mathrmdx + int_1-ln 2^1 1 \, mathrmdx + int_1^e^3 0 \, mathrmdx$$I = \int_{0}^{1-\ln 2} 2 \, \mathrm{d}x + \int_{1-\ln 2}^{1} 1 \, \mathrm{d}x + \int_{1}^{e^3} 0 \, \mathrm{d}x$$ ### Step 3: Perform Integrations I = 2[x]_0^1-ln 2 + 1[x]_1-ln 2^1 + 0$$I = 2[x]_0^{1-\ln 2} + 1[x]_{1-\ln 2}^{1} + 0$$ I = 2(1 - ln 2 - 0) + 1(1 - (1 - ln 2)) = 2 - 2ln 2 + ln 2 = 2 - ln 2$$I = 2(1 - \ln 2 - 0) + 1(1 - (1 - \ln 2)) = 2 - 2\ln 2 + \ln 2 = 2 - \ln 2$$ ### Step 4: Solve for Alpha Cubed Compare the integrated value to alpha - ln 2$\alpha - \ln 2$: alpha - ln 2 = 2 - ln 2 implies alpha = 2$$\alpha - \ln 2 = 2 - \ln 2 \implies \alpha = 2$$ alpha^3 = 2^3 = 8$$\alpha^3 = 2^3 = 8$$ ### Pattern Recognition Always map the function values at the extreme boundary points first. Tracking the downward path from 2.71 rightarrow 2 rightarrow 1 rightarrow 0$2.71 \rightarrow 2 \rightarrow 1 \rightarrow 0$ reveals exactly where the integer thresholds are crossed. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integrals

### Related Formula Using the properties of definite integrals: int_a^b f(x) \, dx = int_a^b f(a+b-x) \, dx$$\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$$ Also, if f(2a-x) = f(x)$f(2a-x) = f(x)$, then: int_0^2a f(x) \, dx = 2 int_0^a f(x) \, dx$$\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$$ ### Core Logic Let: I = int_0^pi frac8x \, dx4cos^2 x + sin^2 x quad text--- (1)$$I = \int_0^\pi \frac{8x \, dx}{4\cos^2 x + \sin^2 x} \quad \text{--- (1)}$$ Applying x to pi-x$x \to \pi-x$: I = int_0^pi frac8(pi - x) \, dx4cos^2 x + sin^2 x quad text--- (2)$$I = \int_0^\pi \frac{8(\pi - x) \, dx}{4\cos^2 x + \sin^2 x} \quad \text{--- (2)}$$ ### Step 1: Eliminating the x$x$ term in numerator Adding (1) and (2): 2I = 8pi int_0^pi fracdx4cos^2 x + sin^2 x$$2I = 8\pi \int_0^\pi \frac{dx}{4\cos^2 x + \sin^2 x}$$ I = 4pi int_0^pi fracdx4cos^2 x + sin^2 x$$I = 4\pi \int_0^\pi \frac{dx}{4\cos^2 x + \sin^2 x}$$ Since the integrand is symmetric about x = pi/2$x = \pi/2$: I = 8pi int_0^pi/2 fracdx4cos^2 x + sin^2 x$$I = 8\pi \int_0^{\pi/2} \frac{dx}{4\cos^2 x + \sin^2 x}$$ ### Step 2: Integration using sec^2 x$\sec^2 x$ substitution Divide numerator and denominator by cos^2 x$\cos^2 x$: I = 8pi int_0^pi/2 fracsec^2 x \, dx4 + tan^2 x$$I = 8\pi \int_0^{\pi/2} \frac{\sec^2 x \, dx}{4 + \tan^2 x}$$ Let t = tan x implies dt = sec^2 x \, dx$t = \tan x \implies dt = \sec^2 x \, dx$ At x = 0$x = 0$, t = 0$t = 0$; at x = pi/2$x = \pi/2$, t to infty$t \to \infty$. I = 8pi int_0^infty fracdtt^2 + 2^2$$I = 8\pi \int_0^\infty \frac{dt}{t^2 + 2^2}$$ I = 8pi left[ frac12 tan^-1left(fract2right) right]_0^infty = 4pi left( fracpi2 - 0 right) = 2pi^2$$I = 8\pi \left[ \frac{1}{2} \tan^{-1}\left(\frac{t}{2}\right) \right]_0^\infty = 4\pi \left( \frac{pi}{2} - 0 \right) = 2\pi^2$$ ### Pattern Recognition The presence of a linear x$x$ factor in the numerator of a definite integral with symmetric trigonometric bounds is almost always eliminated using the a+b-x$a+b-x$ property. This reduces the integral to a standard substitution problem. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integrals

### 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