### 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

### Related Formula textStandard Integral: int sqrta^2-y^2 dy = fracy2sqrta^2-y^2 + fraca^22sin^-1left(fracyaright) + C$\text{Standard Integral: } \int \sqrt{a^2-y^2} dy = \frac{y}{2}\sqrt{a^2-y^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{y}{a}\right) + C$ textArea of right triangle = frac12 times textbase times textheight$\text{Area of right triangle } = \frac{1}{2} \times \text{base} \times \text{height}$ ### Core Logic First, find the points of intersection between the circle x^2+y^2=169$x^2+y^2=169$ and the line 5x-y=13 Rightarrow y = 5x-13$5x-y=13 \Rightarrow y = 5x-13$. Substitute y$y$ into the circle equation: x^2 + (5x-13)^2 = 169$x^2 + (5x-13)^2 = 169$ x^2 + 25x^2 - 130x + 169 = 169$x^2 + 25x^2 - 130x + 169 = 169$ 26x^2 - 130x = 0 Rightarrow 26x(x - 5) = 0$26x^2 - 130x = 0 \Rightarrow 26x(x - 5) = 0$ The solutions are x=0$x=0$ and x=5$x=5$. When x=0, y=-13$x=0, y=-13$. Point is (0, -13)$(0, -13)$. When x=5, y=12$x=5, y=12$. Point is (5, 12)$(5, 12)$. The required area is bounded below the line x = fracy+135$x = \frac{y+13}{5}$ and above the right-hand boundary of the circle x = sqrt169-y^2$x = \sqrt{169-y^2}$ across the y-axis boundaries [-13, 12]$[-13, 12]$.

### Step 1: Setup Area Integral Integrate with respect to y$y$ (from left to right curves, bounded horizontally): Area = int_-13^12 left( sqrt169-y^2 - fracy+135 right) dy$Area = \int_{-13}^{12} \left( \sqrt{169-y^2} - \frac{y+13}{5} \right) dy$ Split the integral into two parts: Part A (Circle): int_-13^12 sqrt169-y^2 dy$\int_{-13}^{12} \sqrt{169-y^2} dy$ Part B (Line): int_-13^12 fracy+135 dy$\int_{-13}^{12} \frac{y+13}{5} dy$ ### Step 2: Evaluate Integrals Part A (Circle Integral): = left[ fracy2sqrt169-y^2 + frac1692sin^-1left(fracy13right) right]_-13^12$= \left[ \frac{y}{2}\sqrt{169-y^2} + \frac{169}{2}\sin^{-1}\left(\frac{y}{13}\right) \right]_{-13}^{12}$ Evaluate at upper limit 12: = frac122sqrt169-144 + frac1692sin^-1left(frac1213right) = 6(5) + frac1692sin^-1left(frac1213right) = 30 + frac1692sin^-1left(frac1213right)$= \frac{12}{2}\sqrt{169-144} + \frac{169}{2}\sin^{-1}\left(\frac{12}{13}\right) = 6(5) + \frac{169}{2}\sin^{-1}\left(\frac{12}{13}\right) = 30 + \frac{169}{2}\sin^{-1}\left(\frac{12}{13}\right)$ Evaluate at lower limit -13: = 0 + frac1692sin^-1(-1) = -frac169pi4$= 0 + \frac{169}{2}\sin^{-1}(-1) = -\frac{169\pi}{4}$ Value of Part A = 30 + frac169pi4 + frac1692sin^-1left(frac1213right)$30 + \frac{169\pi}{4} + \frac{169}{2}\sin^{-1}\left(\frac{12}{13}\right)$ Part B (Line Integral - matches the area of the bounded triangle geometric region): = frac110 left[ (y+13)^2 right]_-13^12$= \frac{1}{10} \left[ (y+13)^2 \right]_{-13}^{12}$ = frac110(12+13)^2 - 0 = frac25^210 = frac62510 = frac1252 = 62.5$= \frac{1}{10}(12+13)^2 - 0 = \frac{25^2}{10} = \frac{625}{10} = \frac{125}{2} = 62.5$ ### Step 3: Map to Requested Format Subtract Part B from Part A: Area = frac169pi4 + 30 - frac1252 + frac1692sin^-1left(frac1213right)$Area = \frac{169\pi}{4} + 30 - \frac{125}{2} + \frac{169}{2}\sin^{-1}\left(\frac{12}{13}\right)$ Area = frac169pi4 - frac652 + frac1692sin^-1left(frac1213right)$Area = \frac{169\pi}{4} - \frac{65}{2} + \frac{169}{2}\sin^{-1}\left(\frac{12}{13}\right)$ Comparing this exactly with the given format fracpialpha2beta - frac652 + fracalphabetasin^-1(frac1213)$\frac{\pi\alpha}{2\beta} - \frac{65}{2} + \frac{\alpha}{\beta}\sin^{-1}(\frac{12}{13})$: We see that fracalphabeta = frac1692$\frac{\alpha}{\beta} = \frac{169}{2}$. Since 169 and 2 are coprime, alpha = 169$\alpha = 169$ and beta = 2$\beta = 2$. Calculate alpha + beta$\alpha + \beta$: 169 + 2 = 171$169 + 2 = 171$ ### Pattern Recognition When evaluating line integrals forming a triangle with horizontal bounds, bypass algebraic integration and visually calculate frac12 cdot b cdot h$\frac{1}{2} \cdot b \cdot h$. Here, base=25 along y-axis, height=5 along x-axis, area = 125/2$125/2$. Instantly saves integration time. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Application of Integrals Class 11 Mathematics: Straight Lines