If the area of the region left\(x, y): left| 4 - x ^ 2 right| le y le x ^ 2, y le 4, x ge 0 right\$\left\{(x, y): \left| 4 - x ^ {2} \right| \le y \le x ^ {2}, y \le 4, x \ge 0 \right\}$ is left(frac80sqrt2alpha - betaright), alpha, beta in mathbbN$\left(\frac{80\sqrt{2}}{\alpha} - \beta\right), \alpha, \beta \in \mathbb{N}$, then alpha + beta$\alpha + \beta$ is equal to ________.

### Related Formula Area integrating with respect to y$y$: A = int_y_1^y_2 (x_textright - x_textleft) dy$$A = \int_{y_1}^{y_2} (x_{\text{\right}} - x_{\text{\left}}) dy$$ Standard integral: int frac11+y^2 dy = tan^-1(y)$$\int \frac{1}{1+y^2} dy = \tan^{-1}(y)$$ ### Core Logic The boundary curves are: 1) x = frac11+y^2$x = \frac{1}{1+y^2}$ 2) x = fracy^22$x = \frac{y^2}{2}$ Find intersection points by setting x$x$ equal: frac11+y^2 = fracy^22 implies 2 = y^2(1+y^2) implies y^4 + y^2 - 2 = 0$$\frac{1}{1+y^2} = \frac{y^2}{2} \implies 2 = y^2(1+y^2) \implies y^4 + y^2 - 2 = 0$$ (y^2 + 2)(y^2 - 1) = 0$$(y^2 + 2)(y^2 - 1) = 0$$ Since y$y$ is real, y^2 = 1 implies y = pm 1$y^2 = 1 \implies y = \pm 1$. When y = pm 1$y = \pm 1$, x = frac12$x = \frac{1}{2}$. Intersection points are left(frac12, 1right)$\left(\frac{1}{2}, 1\right)$ and left(frac12, -1right)$\left(\frac{1}{2}, -1\right)$. ### Step 1: Set up and Compute Area Integral Between y = -1$y = -1$ and y = 1$y = 1$, frac11+y^2 ge fracy^22$\frac{1}{1+y^2} \ge \frac{y^2}{2}$. textArea = int-1^1 left( frac11+y^2 - fracy^22 right) dy$$\text{Area} = \int{-1}^{1} \left( \frac{1}{1+y^2} - \frac{y^2}{2} \right) dy$$ Since the integrand is an even function of y$y$: textArea = 2 int_0^1 left( frac11+y^2 - fracy^22 right) dy$$\text{Area} = 2 \int_{0}^{1} \left( \frac{1}{1+y^2} - \frac{y^2}{2} \right) dy$$ textArea = 2 left[ tan^-1(y) - fracy^36 right]_0^1$$\text{Area} = 2 \left[ \tan^{-1}(y) - \frac{y^3}{6} \right]_{0}^{1}$$ textArea = 2 left[ tan^-1(1) - frac16 - (0) right] = 2 left( fracpi4 - frac16 right) = fracpi2 - frac13$$\text{Area} = 2 \left[ \tan^{-1}(1) - \frac{1}{6} - (0) \right] = 2 \left( \frac{\pi}{4} - \frac{1}{6} \right) = \frac{\pi}{2} - \frac{1}{3}$$ ### Pattern Recognition Whenever curves are functions of y^2$y^2$, integrating along the y$y$-axis avoids dealing with messy radical functions (square roots) and naturally accounts for symmetry across the x$x$-axis. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Application of Integrals

### Related Formula textArea = int_a^b (y_textupper - y_textlower) \, dx$$\text{Area} = \int_{a}^{b} (y_{\text{upper}} - y_{\text{lower}}) \, dx$$ ### Core Logic Plotting the boundary lines and tracking intersection points yields a composite geometric region bounding a central area.

### Step 1: Set up the integral pieces The bounded region A$A$ can be conceptualized as a total bounding box/rectangle minus specific external integrals: A = 4 - 2 int_0^1 left[ (3 - x) - left( fracx^2 + 32 right) right] \, dx$$A = 4 - 2 \int_{0}^{1} \left[ (3 - x) - \left( \frac{x^2 + 3}{2} \right) \right] \, dx$$ ### Step 2: Evaluate the Integral A = 4 - 2 left[ 3x - fracx^22 - fracx^36 - frac32x right]_0^1$$A = 4 - 2 \left[ 3x - \frac{x^2}{2} - \frac{x^3}{6} - \frac{3}{2}x \right]_{0}^{1}$$ A = 4 - 2 left[ 3 - frac12 - frac16 - frac32 right] = 4 - 2 left[ frac56 right] = 4 - frac53 = frac73$$A = 4 - 2 \left[ 3 - \frac{1}{2} - \frac{1}{6} - \frac{3}{2} \right] = 4 - 2 \left[ \frac{5}{6} \right] = 4 - \frac{5}{3} = \frac{7}{3}$$ ### Step 3: Calculate 6A 6A = 6 times frac73 = 14$6A = 6 \times \frac{7}{3} = 14$ ### Pattern Recognition When dealing with multiple absolute functions (|x|$|x|$, |x-1|$|x-1|$), check for coordinate mirror symmetry across vertical axes to slash total required calculus computations in half. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Application of Integrals

### Related Formula Area enclosed between two intersecting curves y_1 = f(x)$y_1 = f(x)$ and y_2 = g(x)$y_2 = g(x)$ from boundary limits x = a$x = a$ to x = b$x = b$: textArea = int_a^b (y_textupper - y_textlower) \, dx$$\text{Area} = \int_{a}^{b} (y_{\text{upper}} - y_{\text{lower}}) \, dx$$ ### Core Logic Given the two boundary curves: 1. xy + 4y = 16 implies y(x+4) = 16 implies y = frac16x+4$xy + 4y = 16 \implies y(x+4) = 16 \implies y = \frac{16}{x+4}$ 2. x + y = 6 implies y = 6 - x$x + y = 6 \implies y = 6 - x$ Find the intersection points by equating the two expressions for y$y$: frac16x+4 = 6 - x implies 16 = (6-x)(x+4)$$\frac{16}{x+4} = 6 - x \implies 16 = (6-x)(x+4)$$ 16 = 6x + 24 - x^2 - 4x implies x^2 - 2x - 8 = 0$$16 = 6x + 24 - x^2 - 4x \implies x^2 - 2x - 8 = 0$$ (x-4)(x+2) = 0 implies x = 4, \, x = -2 $$(x-4)(x+2) = 0 \implies x = 4, \, x = -2 $$ ### Step 1: Integral Formulation Between x = -2$x = -2$ and x = 4$x = 4$, the line y = 6 - x$y = 6 - x$ lies above the curve y = frac16x+4$y = \frac{16}{x+4}$. Therefore, the required enclosed area is: textArea = int_-2^4 left( (6-x) - frac16x+4 right) dx $$\text{Area} = \int_{-2}^{4} \left( (6-x) - \frac{16}{x+4} \right) dx $$ textArea = left[ 6x - fracx^22 - 16 ln|x+4| right]_-2^4$$\text{Area} = \left[ 6x - \frac{x^2}{2} - 16 \ln|x+4| \right]_{-2}^{4}$$

### Step 2: Apply Limits and Simplify Substitute the upper limit x = 4$x = 4$: U = 6(4) - frac4^22 - 16 ln|4+4| = 24 - 8 - 16 ln 8 = 16 - 16 ln 8$$U = 6(4) - \frac{4^2}{2} - 16 \ln|4+4| = 24 - 8 - 16 \ln 8 = 16 - 16 \ln 8$$ Substitute the lower limit x = -2$x = -2$: L = 6(-2) - frac(-2)^22 - 16 ln|-2+4| = -12 - 2 - 16 ln 2 = -14 - 16 ln 2$$L = 6(-2) - \frac{(-2)^2}{2} - 16 \ln|-2+4| = -12 - 2 - 16 \ln 2 = -14 - 16 \ln 2$$ Subtract the lower limit value from the upper limit value: textArea = U - L = (16 - 16 ln 8) - (-14 - 16 ln 2)$$\text{Area} = U - L = (16 - 16 \ln 8) - (-14 - 16 \ln 2)$$ textArea = 30 - 16 ln(2^3) + 16 ln 2 = 30 - 48 ln 2 + 16 ln 2$$\text{Area} = 30 - 16 \ln(2^3) + 16 \ln 2 = 30 - 48 \ln 2 + 16 \ln 2$$ textArea = 30 - 32 ln 2 = 30 - 32 log_e2 $$\text{Area} = 30 - 32 \ln 2 = 30 - 32 \log_{e}2 $$ ### Pattern Recognition Sees: Area bounded by a straight line and a shifting rectangular hyperbola. Shortcut: The roots of the difference equation x^2 - 2x - 8 = 0$x^2 - 2x - 8 = 0$ directly give the limits. Always check graph orientations to place the upper linear equation before the curve equation inside the integral bracket to preserve absolute area values. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Application of Integrals Class 11 Mathematics: Conic Sections (Hyperbola)

### Related Formula textArea A = int min\f(x), g(x)\\, dx$$\text{Area } A = \int \min\{f(x), g(x)\}\, dx$$ ### Core Logic Let us find the point of intersection of the curves y = x^2 + 2$y = x^2 + 2$ and y = 2x + 2$y = 2x + 2$: x^2 + 2 = 2x + 2 implies x^2 - 2x = 0 implies x = 0 text or x = 2$$x^2 + 2 = 2x + 2 \implies x^2 - 2x = 0 \implies x = 0 \text{ or } x = 2$$ Evaluating behaviors over boundaries: * For 0 leq x leq 2$0 \leq x \leq 2$: x^2 + 2 leq 2x + 2 implies min = x^2 + 2$x^2 + 2 \leq 2x + 2 \implies \min = x^2 + 2$ * For 2 leq x leq 3$2 \leq x \leq 3$: 2x + 2 leq x^2 + 2 implies min = 2x + 2$2x + 2 \leq x^2 + 2 \implies \min = 2x + 2$ ### Step 1: Integration Resolution Setting up continuous area integral steps: A = int_0^2 (x^2 + 2)\, dx + int_2^3 (2x + 2)\, dx$$A = \int_{0}^{2} (x^2 + 2)\, dx + \int_{2}^{3} (2x + 2)\, dx$$ A = left[ fracx^33 + 2x right]_0^2 + left[ x^2 + 2x right]_2^3$$A = \left[ \frac{x^3}{3} + 2x \right]_{0}^{2} + \left[ x^2 + 2x \right]_{2}^{3}$$ A = left( frac83 + 4 right) + left( (9 + 6) - (4 + 4) right)$$A = \left( \frac{8}{3} + 4 \right) + \left( (9 + 6) - (4 + 4) \right)$$ A = frac203 + (15 - 8) = frac203 + 7 = frac413$$A = \frac{20}{3} + (15 - 8) = \frac{20}{3} + 7 = \frac{41}{3}$$

### Step 2: Scaling the Output Value We need to compute 12A$12A$: 12A = 12 times frac413 = 4 times 41 = 164$$12A = 12 \times \frac{41}{3} = 4 \times 41 = 164$$ ### Pattern Recognition For min/max boundary sets, always compute intersections first to accurately split integration domains into separate regions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Application of Integrals

### Related Formula textArea = int_y_1^y_2 (x_textright - x_textleft) dy$$\text{Area} = \int_{y_1}^{y_2} (x_{\text{right}} - x_{\text{left}}) dy$$ ### Core Logic We need to find the area bounded by the curves in the first quadrant (since y ge 0$y \ge 0$). Note that the first inequality x^2y + 4 ge 0$x^2y + 4 \ge 0$ is trivially satisfied for all x, y ge 0$x, y \ge 0$. So we focus on bounding x$x$ using: Right curve: x = 8 - 4y^2$x = 8 - 4y^2$ Left curve: x = -2y^2$x = -2y^2$ However, there might be constraints where these cross or hit the axes. We must inspect intersections. ### Step 1: Finding Intersection Points Where do the left and right parabolas intersect? 8 - 4y^2 = -2y^2 Rightarrow 2y^2 = 8 Rightarrow y^2 = 4 Rightarrow y = 2$8 - 4y^2 = -2y^2 \Rightarrow 2y^2 = 8 \Rightarrow y^2 = 4 \Rightarrow y = 2$ (Since y ge 0$y \ge 0$). Wait, does x$x$ have boundaries? Since x$x$ can't drop arbitrarily into negative territory if it's restricted by other axes. Let's check x^2y + 4 ge 0$x^2y + 4 \ge 0$. If x = -2y^2$x = -2y^2$, then y$y$ must be bounded, but wait - the question limits are actually split into two regions depending on x^2y+4 ge 0$x^2y+4 \ge 0$? Wait, the first inequality x^2y+4 ge 0$x^2y+4 \ge 0$ might not be trivial if x$x$ is negative. If x = -2y^2$x = -2y^2$, then (-2y^2)^2 y + 4 ge 0 Rightarrow 4y^5 + 4 ge 0$(-2y^2)^2 y + 4 \ge 0 \Rightarrow 4y^5 + 4 \ge 0$, which is true for all y ge 0$y \ge 0$. Wait, there seems to be a misinterpretation of the first inequality. Let's look closer at the PDF solution boundaries. The integration is broken at y=1$y=1$. Why? x+2y^2 ge 0 Rightarrow x ge -2y^2$x+2y^2 \ge 0 \Rightarrow x \ge -2y^2$. Wait, the solution states another left curve: 2y-4$2y-4$. Is x^2y+4 ge 0$x^2y+4 \ge 0$ actually x + 2y - 4 ge 0$x + 2y - 4 \ge 0$? Yes, OCR shows `x^2y+4` but the solution integrates `(2y-4)`. Thus the original condition is likely x - 2y + 4 ge 0 Rightarrow x ge 2y - 4$x - 2y + 4 \ge 0 \Rightarrow x \ge 2y - 4$! Let's assume the left boundary splits between x = -2y^2$x = -2y^2$ and x = 2y - 4$x = 2y - 4$. ### Step 2: Region Bounds Intersection of x = -2y^2$x = -2y^2$ and x = 2y - 4$x = 2y - 4$: -2y^2 = 2y - 4 Rightarrow y^2 + y - 2 = 0 Rightarrow (y+2)(y-1) = 0 Rightarrow y = 1$-2y^2 = 2y - 4 \Rightarrow y^2 + y - 2 = 0 \Rightarrow (y+2)(y-1) = 0 \Rightarrow y = 1$. Intersection of x = 2y - 4$x = 2y - 4$ and x = 8 - 4y^2$x = 8 - 4y^2$: 2y - 4 = 8 - 4y^2 Rightarrow 4y^2 + 2y - 12 = 0 Rightarrow 2y^2 + y - 6 = 0 Rightarrow (2y-3)(y+2) = 0 Rightarrow y = 3/2$2y - 4 = 8 - 4y^2 \Rightarrow 4y^2 + 2y - 12 = 0 \Rightarrow 2y^2 + y - 6 = 0 \Rightarrow (2y-3)(y+2) = 0 \Rightarrow y = 3/2$. So the region shifts left boundary at y=1$y=1$ and closes entirely at y=3/2$y=3/2$. ### Step 3: Setting up the Integration Region 1 (from y=0$y=0$ to y=1$y=1$): A_1 = int_0^1 ((8 - 4y^2) - (-2y^2)) dy = int_0^1 (8 - 2y^2) dy$$A_1 = \int_0^1 ((8 - 4y^2) - (-2y^2)) dy = \int_0^1 (8 - 2y^2) dy$$ A_1 = left[ 8y - frac2y^33 right]_0^1 = 8 - frac23 = frac223$$A_1 = \left[ 8y - \frac{2y^3}{3} \right]_0^1 = 8 - \frac{2}{3} = \frac{22}{3}$$ Region 2 (from y=1$y=1$ to y=3/2$y=3/2$): A_2 = int_1^3/2 ((8 - 4y^2) - (2y - 4)) dy = int_1^3/2 (12 - 2y - 4y^2) dy$$A_2 = \int_1^{3/2} ((8 - 4y^2) - (2y - 4)) dy = \int_1^{3/2} (12 - 2y - 4y^2) dy$$ A_2 = left[ 12y - y^2 - frac4y^33 right]_1^3/2$$A_2 = \left[ 12y - y^2 - \frac{4y^3}{3} \right]_1^{3/2}$$ A_2 = left( 12left(frac32right) - frac94 - frac43left(frac278right) right) - left( 12 - 1 - frac43 right)$$A_2 = \left( 12\left(\frac{3}{2}\right) - \frac{9}{4} - \frac{4}{3}\left(\frac{27}{8}\right) \right) - \left( 12 - 1 - \frac{4}{3} \right)$$ A_2 = left( 18 - frac94 - frac92 right) - left( 11 - frac43 right) = left( 18 - frac274 right) - frac293 = frac454 - frac293 = frac135 - 11612 = frac1912$$A_2 = \left( 18 - \frac{9}{4} - \frac{9}{2} \right) - \left( 11 - \frac{4}{3} \right) = \left( 18 - \frac{27}{4} \right) - \frac{29}{3} = \frac{45}{4} - \frac{29}{3} = \frac{135 - 116}{12} = \frac{19}{12}$$ ### Step 4: Final Output Total Area A = A_1 + A_2$A = A_1 + A_2$: A = frac223 + frac1912 = frac88 + 1912 = frac10712$$A = \frac{22}{3} + \frac{19}{12} = \frac{88 + 19}{12} = \frac{107}{12}$$ This implies m = 107$m = 107$ and n = 12$n = 12$. Since 107 and 12 are coprime, m + n = 107 + 12 = 119$m + n = 107 + 12 = 119$. ### Pattern Recognition When dealing with multiple inequalities bounded by y ge 0$y \ge 0$, always project horizontally (integrate wrt y$y$) as the bounds natively trace left-to-right distances. Find intersection nodes to partition the integral correctly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Application of Integrals