### Core Logic Let the equation of the tangent line passing through A(-2,0)$A(-2,0)$ be: y = m(x + 2) implies x = fracym - 2$y = m(x + 2) \implies x = \frac{y}{m} - 2$ The equation of the parabola is y^2 = x - 2 implies x = y^2 + 2$y^2 = x - 2 \implies x = y^2 + 2$. Substituting x$x$ from the line into the parabola: y^2 + 2 = fracym - 2 implies y^2 - fracym + 4 = 0$y^2 + 2 = \frac{y}{m} - 2 \implies y^2 - \frac{y}{m} + 4 = 0$ For the line to be a tangent, the discriminant of this quadratic equation must be zero (D = 0$D = 0$): left(-frac1mright)^2 - 4(1)(4) = 0 implies frac1m^2 = 16 implies m = pm frac14$\left(-\frac{1}{m}\right)^2 - 4(1)(4) = 0 \implies \frac{1}{m^2} = 16 \implies m = \pm \frac{1}{4}$ Since point B$B$ is in the first quadrant, the slope must be positive, so m = frac14$m = \frac{1}{4}$. The line equation is y = frac14(x + 2) implies x = 4y - 2$y = \frac{1}{4}(x + 2) \implies x = 4y - 2$. The point of tangency B$B$ is found at y = frac12m = 2$y = \frac{1}{2m} = 2$, which gives x = 6$x = 6$, so B = (6,2)$B = (6,2)$. ### Step 1: Setting up the Area Integral Integrating with respect to y$y$ avoids splitting the region into two parts along the x-axis: textArea = int_0^2 left(x_textparabola - x_textlineright) dy$\text{Area} = \int_{0}^{2} \left(x_{\text{parabola}} - x_{\text{line}}\right) dy$ textArea = int_0^2 left((y^2 + 2) - (4y - 2)right) dy = int_0^2 left(y^2 - 4y + 4right) dy$\text{Area} = \int_{0}^{2} \left((y^2 + 2) - (4y - 2)\right) dy = \int_{0}^{2} \left(y^2 - 4y + 4\right) dy$

### Step 2: Evaluating the Integral Integrating term by term: textArea = left[ fracy^33 - 2y^2 + 4y right]_0^2$\text{Area} = \left[ \frac{y^3}{3} - 2y^2 + 4y \right]_{0}^{2}$ textArea = left( frac83 - 2(4) + 4(2) right) - 0 = frac83 - 8 + 8 = frac83$\text{Area} = \left( \frac{8}{3} - 2(4) + 4(2) \right) - 0 = \frac{8}{3} - 8 + 8 = \frac{8}{3}$ ### Pattern Recognition Integrating with respect to y$y$ (horizontal strips) when dealing with horizontal parabolas or lines crossing the x-axis eliminates the need to break your area computation into multiple piecewise integrals. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Area Under Curves Class 11 Mathematics: Conic Sections

### Related Formula Area tracking equation via horizontal slices or split verticals: textArea = int_x_1^x_2 (y_textupper - y_textlower)\,mathrmdx$\text{Area} = \int_{x_1}^{x_2} (y_{\text{upper}} - y_{\text{lower}})\,\mathrm{d}x$ ### Core Logic Find the intersections of y = |x - 5|$y = |x - 5|$ and y = 4sqrtx$y = 4\sqrt{x}$: Branch 1: 5 - x = 4sqrtx implies x + 4sqrtx - 5 = 0 implies (sqrtx + 5)(sqrtx - 1) = 0 implies x = 1, y = 4$5 - x = 4\sqrt{x} \implies x + 4\sqrt{x} - 5 = 0 \implies (\sqrt{x} + 5)(\sqrt{x} - 1) = 0 \implies x = 1, y = 4$. Branch 2: x - 5 = 4sqrtx implies x - 4sqrtx - 5 = 0 implies (sqrtx - 5)(sqrtx + 1) = 0 implies x = 25, y = 20$x - 5 = 4\sqrt{x} \implies x - 4\sqrt{x} - 5 = 0 \implies (\sqrt{x} - 5)(\sqrt{x} + 1) = 0 \implies x = 25, y = 20$.

### Step 1: Set Up Area Definite Integral Split integration intervals around vertex x = 5$x = 5$ or compute via simple boundary differences: A = int_1^25 4sqrtx\,mathrmdx - textArea of Left Triangle - textArea of Right Triangle$A = \int_{1}^{25} 4\sqrt{x}\,\mathrm{d}x - \text{Area of Left Triangle} - \text{Area of Right Triangle}$ textArea of Left Triangle = frac12 times (5 - 1) times 4 = 8$\text{Area of Left Triangle} = \frac{1}{2} \times (5 - 1) \times 4 = 8$ textArea of Right Triangle = frac12 times (25 - 5) times 20 = 200$\text{Area of Right Triangle} = \frac{1}{2} \times (25 - 5) \times 20 = 200$ ### Step 2: Complete Computations int_1^25 4sqrtx\,mathrmdx = left[ frac83x^3/2 right]_1^25 = frac83(125 - 1) = frac8 times 1243 = frac9923$\int_{1}^{25} 4\sqrt{x}\,\mathrm{d}x = \left[ \frac{8}{3}x^{3/2} \right]_{1}^{25} = \frac{8}{3}(125 - 1) = \frac{8 \times 124}{3} = \frac{992}{3}$ A = frac9923 - 208 = frac992 - 6243 = frac3683$A = \frac{992}{3} - 208 = \frac{992 - 624}{3} = \frac{368}{3}$ 3A = 368$3A = 368$ ### Pattern Recognition Subtracting standard geometric triangles beneath linear configurations from total absolute root curves saves significant time compared to managing multiple separate analytical integral pieces. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Area Under Curves

### Related Formula The area between two curves y_1(x)$y_1(x)$ and y_2(x)$y_2(x)$ from x = a$x = a$ to x = b$x = b$ is given by: textArea = int_a^b |y_1(x) - y_2(x)| dx$\text{Area} = \int_{a}^{b} |y_1(x) - y_2(x)| dx$ ### Core Logic Analyze the curves to locate intersection points : - Curve 1: y = e^x$y = e^x$ - Curve 2: y = |e^x - 1|$y = |e^x - 1|$ - Boundary: y-axis (x = 0$x = 0$) For x < 0$x < 0$, e^x < 1 Rightarrow |e^x - 1| = 1 - e^x$e^x < 1 \Rightarrow |e^x - 1| = 1 - e^x$ . Find intersection: e^x = 1 - e^x Rightarrow 2e^x = 1 Rightarrow e^x = frac12 Rightarrow x = -ln 2$e^x = 1 - e^x \Rightarrow 2e^x = 1 \Rightarrow e^x = \frac{1}{2} \Rightarrow x = -\ln 2$.

### Step 1: Set up the Definite Integral The integration spans from x = -ln 2$x = -\ln 2$ to x = 0$x = 0$. In this interval, e^x ge 1 - e^x$e^x \ge 1 - e^x$: textArea = int_-ln 2^0 left[ e^x - (1 - e^x) right] dx$\text{Area} = \int_{-\ln 2}^{0} \left[ e^x - (1 - e^x) \right] dx$ textArea = int_-ln 2^0 (2e^x - 1) dx$\text{Area} = \int_{-\ln 2}^{0} (2e^x - 1) dx$ ### Step 2: Integration Evaluation Integrate term-by-term : textArea = left[ 2e^x - x right]_-ln 2^0$\text{Area} = \left[ 2e^x - x \right]_{-\ln 2}^{0}$ = left(2e^0 - 0right) - left(2e^-ln 2 - (-ln 2)right)$= \left(2e^0 - 0\right) - \left(2e^{-\ln 2} - (-\ln 2)\right)$ = 2 - left(2left(frac12right) + ln 2right) = 2 - (1 + ln 2) = 1 - ln 2$= 2 - \left(2\left(\frac{1}{2}\right) + \ln 2\right) = 2 - (1 + \ln 2) = 1 - \ln 2$ ### Pattern Recognition Modulus graphs split at their critical point (e^x = 1 Rightarrow x = 0$e^x = 1 \Rightarrow x = 0$). Recognizing that the required segment lies strictly in the negative quadrant simplifies the definition of the upper and lower functions immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Application of Integrals

### Related Formula The net area between two intersecting functions y_2(x)$y_2(x)$ and y_1(x)$y_1(x)$ from lower limit a$a$ to upper limit b$b$ is evaluated using the definite integral: textArea = int_a^b [y_2(x) - y_1(x)] dx$\text{Area} = \int_{a}^{b} [y_2(x) - y_1(x)] dx$ ### Core Logic Let's perform a horizontal shift transformation by defining X = x + 2$X = x + 2$ to simplify the expressions: - Parabola: y = x^2 + 4x + 2 = (x+2)^2 - 2 implies y = X^2 - 2$y = x^2 + 4x + 2 = (x+2)^2 - 2 \implies y = X^2 - 2$ - Absolute curve: y = |x + 2| implies y = |X|$y = |x + 2| \implies y = |X|$ This coordinate translation preserves area completely while shifting the axis to the origin. ### Step 1: Compute Points of Intersection Find where the transformed curves intersect by equating the functions: |X|^2 - 2 = |X|$|X|^2 - 2 = |X|$ |X|^2 - |X| - 2 = 0$|X|^2 - |X| - 2 = 0$ (|X| - 2)(|X| + 1) = 0$(|X| - 2)(|X| + 1) = 0$ Since |X| geq 0$|X| \geq 0$, we discard |X| = -1$|X| = -1$. This gives |X| = 2 implies X = pm 2$|X| = 2 \implies X = \pm 2$. ### Step 2: Set up and Evaluate the Transformed Integral Since both curves are symmetric about the vertical line X = 0$X = 0$, we can compute the area for the positive half and double it: textArea = 2 int_0^2 left[ X - (X^2 - 2) right] dX = 2 int_0^2 left( 2 + X - X^2 right) dX$\text{Area} = 2 \int_{0}^{2} \left[ X - (X^2 - 2) \right] dX = 2 \int_{0}^{2} \left( 2 + X - X^2 \right) dX$ Integrate the polynomial row-by-row: = 2 left[ 2X + fracX^22 - fracX^33 right]_0^2$= 2 \left[ 2X + \frac{X^2}{2} - \frac{X^3}{3} \right]_{0}^{2}$ = 2 left[ 2(2) + frac42 - frac83 right] = 2 left[ 4 + 2 - frac83 right]$= 2 \left[ 2(2) + \frac{4}{2} - \frac{8}{3} \right] = 2 \left[ 4 + 2 - \frac{8}{3} \right]$ = 2 left[ 6 - frac83 right] = 2 cdot frac103 = frac203$= 2 \left[ 6 - \frac{8}{3} \right] = 2 \cdot \frac{10}{3} = \frac{20}{3}$ ### Pattern Recognition Applying a horizontal variable substitution matching X = x + ± c$X = x + ± c$ shifts complex quadratic forms to centered formats, which avoids tedious arithmetic across asymmetrical integration limits. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Application of Integrals

### 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