Let the area of the bounded region (x,y):0leq 9xleq y^2,ygeq 3x - 6${(x,y):0\leq 9x\leq y^2,y\geq 3x - 6}$ be A. Then 6A is equal to

### 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

### Related Formula textArea = int_y_1^y_2 (x_R - x_L) dy$$\text{Area} = \int_{y_1}^{y_2} (x_R - x_L) dy$$ ### Core Logic

To simplify calculations, shift the origin. Let X = x - 2$X = x - 2$. The equations become: Parabola: y^2 = 4X Rightarrow X = fracy^24$y^2 = 4X \Rightarrow X = \frac{y^2}{4}$ Line: y = 2(X + 2) - 8 Rightarrow y = 2X - 4 Rightarrow X = fracy + 42$y = 2(X + 2) - 8 \Rightarrow y = 2X - 4 \Rightarrow X = \frac{y + 4}{2}$ ### Step 1: Finding points of intersection Set the X$X$ values equal to find intersection points in terms of y$y$: fracy^24 = fracy + 42$$\frac{y^2}{4} = \frac{y + 4}{2}$$ y^2 = 2y + 8$y^2 = 2y + 8$ y^2 - 2y - 8 = 0$$y^2 - 2y - 8 = 0$$ (y - 4)(y + 2) = 0$$(y - 4)(y + 2) = 0$$ The intersection points are at y = -2$y = -2$ and y = 4$y = 4$. ### Step 2: Area Integration Integrate with respect to y$y$ from -2$-2$ to 4$4$: A = int_-2^4 left( x_R - x_L right) dy$$A = \int_{-2}^{4} \left( x_R - x_L \right) dy$$ A = int_-2^4 left( fracy + 42 - fracy^24 right) dy$$A = \int_{-2}^{4} \left( \frac{y + 4}{2} - \frac{y^2}{4} \right) dy$$ A = left[ fracy^24 + 2y - fracy^312 right]_-2^4$$A = \left[ \frac{y^2}{4} + 2y - \frac{y^3}{12} \right]_{-2}^{4}$$ Upper limit (y=4)$(y=4)$: frac164 + 8 - frac6412 = 4 + 8 - frac163 = 12 - frac163 = frac203$\frac{16}{4} + 8 - \frac{64}{12} = 4 + 8 - \frac{16}{3} = 12 - \frac{16}{3} = \frac{20}{3}$ Lower limit (y=-2)$(y=-2)$: frac44 - 4 - frac-812 = 1 - 4 + frac23 = -3 + frac23 = -frac73$\frac{4}{4} - 4 - \frac{-8}{12} = 1 - 4 + \frac{2}{3} = -3 + \frac{2}{3} = -\frac{7}{3}$ A = frac203 - left(-frac73right) = frac273 = 9$$A = \frac{20}{3} - \left(-\frac{7}{3}\right) = \frac{27}{3} = 9$$ The solution simplifies it directly to 9$9$ square units. ### Pattern Recognition For a horizontal parabola interacting with a line, integrating along the y-axis is always cleaner than splitting it into multiple integrals along the x-axis. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: 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

Find intersection points of y = 4x - x^2$y = 4x - x^2$ and 3y = (x - 4)^2$3y = (x - 4)^2$: 3(4x - x^2) = x^2 - 8x + 16$$3(4x - x^2) = x^2 - 8x + 16$$ 12x - 3x^2 = x^2 - 8x + 16$$12x - 3x^2 = x^2 - 8x + 16$$ 4x^2 - 20x + 16 = 0 implies x^2 - 5x + 4 = 0$$4x^2 - 20x + 16 = 0 \implies x^2 - 5x + 4 = 0$$ Roots are x = 1, 4$x = 1, 4$. Area integral: textArea = int_1^4 left[ (4x - x^2) - frac(x - 4)^23 right] dx$$\text{Area} = \int_1^4 \left[ (4x - x^2) - \frac{(x - 4)^2}{3} \right] dx$$ = left[ frac4x^22 - fracx^33 - frac(x - 4)^39 right]_1^4$$= \left[ \frac{4x^2}{2} - \frac{x^3}{3} - \frac{(x - 4)^3}{9} \right]_1^4$$ = left[ 2(16) - frac643 - 0 right] - left[ 2(1) - frac13 - frac(-3)^39 right]$$= \left[ 2(16) - \frac{64}{3} - 0 \right] - \left[ 2(1) - \frac{1}{3} - \frac{(-3)^3}{9} \right]$$ = left( 32 - frac643 right) - left( 2 - frac13 + 3 right)$$= \left( 32 - \frac{64}{3} \right) - \left( 2 - \frac{1}{3} + 3 \right)$$ = frac323 - left( 5 - frac13 right) = frac323 - frac143 = frac183 = 6$$= \frac{32}{3} - \left( 5 - \frac{1}{3} \right) = \frac{32}{3} - \frac{14}{3} = \frac{18}{3} = 6$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Applications of the Integrals

### Core Logic Given y^2 le 4x$y^2 \le 4x$ and x < 4$x < 4$. Analyze the inequality fracxy(x-1)(x-2)(x-3)(x-4) > 0$\frac{xy(x-1)(x-2)}{(x-3)(x-4)} > 0$ considering y > 0$y > 0$ and y < 0$y < 0$ separately.

### Step 1: Case I (y > 0) If y > 0$y > 0$, the inequality reduces to fracx(x-1)(x-2)(x-3)(x-4) > 0$\frac{x(x-1)(x-2)}{(x-3)(x-4)} > 0$. Using wavy curve method and given x in (0, 4)$x \in (0, 4)$: x in (0, 1) cup (2, 3)$x \in (0, 1) \cup (2, 3)$. ### Step 2: Case II (y < 0) If y < 0$y < 0$, the inequality reduces to fracx(x-1)(x-2)(x-3)(x-4) < 0$\frac{x(x-1)(x-2)}{(x-3)(x-4)} < 0$. Using wavy curve method and given x in (0, 4)$x \in (0, 4)$: x in (1, 2) cup (3, 4)$x \in (1, 2) \cup (3, 4)$. ### Step 3: Area Computation Because the regions map perfectly without overlap in opposite quadrants relative to the x-axis, they form complete parabolic strips when combined: Area = 2 int_0^4 sqrtx dx = 2 cdot frac23[x^3/2]_0^4 = frac43 cdot 8 = frac323$= 2 \int_{0}^{4} \sqrt{x} dx = 2 \cdot \frac{2}{3}[x^{3/2}]_{0}^{4} = \frac{4}{3} \cdot 8 = \frac{32}{3}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Area Under Curves Class 11 Maths: Linear Inequalities