Let the area of the region bounded by the curve y = max \sin x, cos x\$y = \max \{\sin x, \cos x\}$, lines x = 0, x = frac3pi2$x = 0, x = \frac{3\pi}{2}$, and the x$x$-axis be A$A$. Then, A + A^2$A + A^2$ is equal to _____.

### Related Formula Area of an ellipse fracx^2a^2 + fracy^2b^2 = 1$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by: textArea = pi a b$$\text{Area} = \pi a b$$ Area of a rhombus bounded by |x| + |y| = a$|x| + |y| = a$ is 2a^2$2a^2$. ### Core Logic The given curves form a bounded geometric area. Ellipse: x^2 + 4y^2 = 4 Rightarrow fracx^24 + fracy^21 = 1$x^2 + 4y^2 = 4 \Rightarrow \frac{x^2}{4} + \frac{y^2}{1} = 1$. Here, a = 2$a = 2$, b = 1$b = 1$. The region to be excluded is bounded by y = |x| - 1$y = |x| - 1$ and y = 1 - |x|$y = 1 - |x|$, which rearranges to |x| + |y| = 1$|x| + |y| = 1$. This forms a square/rhombus centered at the origin with vertices at (1, 0), (0, 1), (-1, 0), (0, -1)$(1, 0), (0, 1), (-1, 0), (0, -1)$. ### Step 1: Calculate Total and Excluded Areas Total Area of the Ellipse: textArea = pi (2)(1) = 2pi$$\text{Area} = \pi (2)(1) = 2\pi$$ Excluded Area (Rhombus |x| + |y| = 1$|x| + |y| = 1$): The rhombus consists of 4 identical right-angled triangles in each quadrant. Area of one triangle = frac12 times textbase times textheight = frac12 times 1 times 1 = frac12$\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$. Total excluded area = 4 times frac12 = 2$4 \times \frac{1}{2} = 2$. ### Step 2: Calculate Required Area

Required Area = Area of ellipse -$-$ Shaded Area = 2pi - 2$= 2\pi - 2$ = 2(pi - 1)$= 2(\pi - 1)$ ### Pattern Recognition Transform absolute value equations y = pm(|x| - a)$y = \pm(|x| - a)$ into |x| + |y| = a$|x| + |y| = a$ to instantly recognize a standard rhombus, allowing direct geometry formulas instead of integration. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Area Under Curves Class 11 Maths: Conic Sections

### Related Formula textArea = int_x_1^x_2 (f(x) - g(x)) dx$$\text{Area} = \int_{x_1}^{x_2} (f(x) - g(x)) dx$$ textArea of Parabola piece: int_0^2 (4-x^2) dx$$\text{Area of Parabola piece: } \int_{0}^{2} (4-x^2) dx$$ ### Core Logic

The region is bounded above by y = 4 - x^2$y = 4 - x^2$, below by y = 1 - 2x$y = 1 - 2x$, and constrained to x geq 0, y geq 0$x \geq 0, y \geq 0$. The parabola intersects the x-axis at x=2$x=2$ (since 4-x^2=0, xgeq 0$4-x^2=0, x\geq 0$). The line intersects the x-axis at x=frac12$x=\frac{1}{2}$ (since 1-2x=0$1-2x=0$) and y-axis at y=1$y=1$. The required area is the area under the parabola in the first quadrant minus the small triangular region bounded by y = 1-2x, x=0, y=0$y = 1-2x, x=0, y=0$. ### Step 1: Calculate the Area textTotal area under parabola in 1st quadrant = int_0^2 (4 - x^2) dx = left[ 4x - fracx^33 right]_0^2 = 8 - frac83 = frac163$$\text{Total area under parabola in 1st quadrant} = \int_0^2 (4 - x^2) dx = \left[ 4x - \frac{x^3}{3} \right]_0^2 = 8 - \frac{8}{3} = \frac{16}{3}$$ Area of the small right triangle formed by the line y=1-2x$y=1-2x$ in the first quadrant: Vertices are (0,0), (1/2,0), (0,1)$(0,0), (1/2,0), (0,1)$. textArea of triangle = frac12 times textbase times textheight = frac12 times frac12 times 1 = frac14$$\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{1}{2} \times 1 = \frac{1}{4}$$ ### Step 2: Subtraction and Format Match textRequired Area = frac163 - frac14 = frac64 - 312 = frac6112$$\text{Required Area} = \frac{16}{3} - \frac{1}{4} = \frac{64 - 3}{12} = \frac{61}{12}$$ Here, alpha = 61, beta = 12$\alpha = 61, \beta = 12$. Check gcd(61, 12) = 1$\gcd(61, 12) = 1$. This matches. So, alpha + beta = 61 + 12 = 73$\alpha + \beta = 61 + 12 = 73$. ### Pattern Recognition For areas defined by y geq g(x)$y \geq g(x)$ when g(x)$g(x)$ forms a simple geometric shape (like a line), subtract the geometric area directly rather than splitting the integral algebraically. It eliminates integration errors. ### 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 First, find the points of intersection for the curves y = 1 + x^2$y = 1 + x^2$ and y = 3 - x$y = 3 - x$: 1 + x^2 = 3 - x$1 + x^2 = 3 - x$ x^2 + x - 2 = 0$x^2 + x - 2 = 0$ (x + 2)(x - 1) = 0 implies x = -2, x = 1$$(x + 2)(x - 1) = 0 \implies x = -2, x = 1$$ So the total region is bounded between x = -2$x = -2$ and x = 1$x = 1$. The line x = -1$x = -1$ divides this region into two parts.

### Step 1: Setting up the Areas Let the area to the left of x = -1$x = -1$ be proportional to m$m$ and the area to the right be proportional to n$n$. A_m = int_-2^-1 [(3 - x) - (1 + x^2)] \,dx$$A_m = \int_{-2}^{-1} [(3 - x) - (1 + x^2)] \,dx$$ A_n = int_-1^1 [(3 - x) - (1 + x^2)] \,dx$$A_n = \int_{-1}^{1} [(3 - x) - (1 + x^2)] \,dx$$ The integrand simplifies to 2 - x - x^2$2 - x - x^2$.

### Step 2: Integration int (2 - x - x^2) \,dx = 2x - fracx^22 - fracx^33$$\int (2 - x - x^2) \,dx = 2x - \frac{x^2}{2} - \frac{x^3}{3}$$ Evaluate A_m$A_m$: A_m = left[2x - fracx^22 - fracx^33right]_-2^-1$$A_m = \left[2x - \frac{x^2}{2} - \frac{x^3}{3}\right]_{-2}^{-1}$$ A_m = left(-2 - frac12 + frac13right) - left(-4 - 2 + frac83right) = left(-frac136right) - left(-frac103right) = frac76$$A_m = \left(-2 - \frac{1}{2} + \frac{1}{3}\right) - \left(-4 - 2 + \frac{8}{3}\right) = \left(-\frac{13}{6}\right) - \left(-\frac{10}{3}\right) = \frac{7}{6}$$ Evaluate A_n$A_n$: A_n = left[2x - fracx^22 - fracx^33right]_-1^1$$A_n = \left[2x - \frac{x^2}{2} - \frac{x^3}{3}\right]_{-1}^{1}$$ A_n = left(2 - frac12 - frac13right) - left(-2 - frac12 + frac13right) = left(frac76right) - left(-frac136right) = frac206$$A_n = \left(2 - \frac{1}{2} - \frac{1}{3}\right) - \left(-2 - \frac{1}{2} + \frac{1}{3}\right) = \left(\frac{7}{6}\right) - \left(-\frac{13}{6}\right) = \frac{20}{6}$$ ### Step 3: Finding the Ratio The ratio of the areas m:n$m:n$ is: fracmn = fracA_nA_m text or fracA_mA_n$$\frac{m}{n} = \frac{A_n}{A_m} \text{ or } \frac{A_m}{A_n}$$ Wait, the solution designates fracmn = fracint_-1^1int_-2^-1 = frac20/67/6 = frac207$\frac{m}{n} = \frac{\int_{-1}^{1}}{\int_{-2}^{-1}} = \frac{20/6}{7/6} = \frac{20}{7}$. (Since gcd(20,7)=1$\gcd(20,7)=1$, m=20$m=20$ and n=7$n=7$). Therefore, m+n = 20 + 7 = 27$m+n = 20 + 7 = 27$. ### Pattern Recognition When a vertical line divides an area into a ratio, calculate the definite integral on both sides of the splitting line independently. Keep fractions with a common denominator until the final ratio step. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Applications of Integrals

### Related Formula int sqrta^2 - x^2 dx = fracx2sqrta^2 - x^2 + fraca^22sin^-1left(fracxaright)$$\int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$$ ### Core Logic

The intersection points of x^2 + y^2 = 4$x^2 + y^2 = 4$ and x^2 + (y-2)^2 = 4$x^2 + (y-2)^2 = 4$: Subtracting the equations gives y^2 - (y-2)^2 = 0 implies y^2 - (y^2 - 4y + 4) = 0 implies 4y = 4 implies y = 1$y^2 - (y-2)^2 = 0 \implies y^2 - (y^2 - 4y + 4) = 0 \implies 4y = 4 \implies y = 1$. Substitute y=1$y=1$ back: x^2 + 1 = 4 implies x = pm sqrt3$x^2 + 1 = 4 \implies x = \pm \sqrt{3}$. The points of intersection are (sqrt3, 1)$(\sqrt{3}, 1)$ and (-sqrt3, 1)$(-\sqrt{3}, 1)$. The area is symmetric about the y-axis, so we integrate from x = 0$x = 0$ to x = sqrt3$x = \sqrt{3}$ and double the result. Upper curve is the lower arc of the top circle: y = 2 - sqrt4-x^2$y = 2 - \sqrt{4-x^2}$ -- wait, the area is bounded by the top arc of the bottom circle and the bottom arc of the top circle. Upper curve for area: y = sqrt4-x^2$y = \sqrt{4-x^2}$ (from x^2+y^2=4$x^2+y^2=4$) Lower curve for area: y = 2 - sqrt4-x^2$y = 2 - \sqrt{4-x^2}$ (from x^2+(y-2)^2=4$x^2+(y-2)^2=4$) A = 2int_0^sqrt3 left[sqrt4-x^2 - (2 - sqrt4-x^2)right] dx$$A = 2\int_{0}^{\sqrt{3}} \left[\sqrt{4-x^2} - (2 - \sqrt{4-x^2})\right] dx$$ ### Step 1: Integration A = 2int_0^sqrt3 (2sqrt4-x^2 - 2) dx = 4int_0^sqrt3 (sqrt4-x^2 - 1) dx$$A = 2\int_{0}^{\sqrt{3}} (2\sqrt{4-x^2} - 2) dx = 4\int_{0}^{\sqrt{3}} (\sqrt{4-x^2} - 1) dx$$ A = 4left[ frac12left(xsqrt4-x^2 + 4sin^-1fracx2right) - x right]_0^sqrt3$$A = 4\left[ \frac{1}{2}\left(x\sqrt{4-x^2} + 4\sin^{-1}\frac{x}{2}\right) - x \right]_{0}^{\sqrt{3}}$$ Evaluating the limits: = 4left[ frac12left(sqrt3(1) + 4sin^-1left(fracsqrt32right)right) - sqrt3 - (0) right]$$= 4\left[ \frac{1}{2}\left(\sqrt{3}(1) + 4\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)\right) - \sqrt{3} - (0) \right]$$ = 4left[ frac12left(sqrt3 + 4left(fracpi3right)right) - sqrt3 right]$$= 4\left[ \frac{1}{2}\left(\sqrt{3} + 4\left(\frac{\pi}{3}\right)\right) - \sqrt{3} \right]$$ = 4left[ fracsqrt32 + frac2pi3 - sqrt3 right]$$= 4\left[ \frac{\sqrt{3}}{2} + \frac{2\pi}{3} - \sqrt{3} \right]$$ = 4left[ frac2pi3 - fracsqrt32 right]$$= 4\left[ \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \right]$$ = frac8pi3 - 2sqrt3 = frac23(4pi - 3sqrt3) text (Sq. units)$$= \frac{8\pi}{3} - 2\sqrt{3} = \frac{2}{3}(4\pi - 3\sqrt{3}) \text{ (Sq. units)}$$ ### Pattern Recognition For intersecting identical circles with centres on an axis, symmetry simplifies the integration drastically. Recognize that integrating the circle arc function handles the bulk of the calculation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Area Under Curves