### 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