Let A₁$A_{1}$ be the bounded area enclosed by the curves y = x² + 2$y = x^{2} + 2$, x + y = 8$x + y = 8$ and y-axis that lies in the first quadrant. Let A₂$A_{2}$ be the bounded area enclosed by the curves y = x² + 2$y = x^{2} + 2$, y² = x$y^{2} = x$, x = 2$x = 2$, and y-axis that lies in the first quadrant. Then A₁ - A₂$A_{1} - A_{2}$ is equal to

Related Formula

Core Logic

Find intersection points of y = 4x - x²$y = 4x - x^2$ and 3y = (x - 4)²$3y = (x - 4)^2$:

Roots are x = 1, 4$x = 1, 4$.

Area integral:

Chapter Mix

Class 12 Maths: Applications of the Integrals

Core Logic

Given y² ≤ 4x$y^2 \le 4x$ and x < 4$x < 4$. Analyze the inequality (xy(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 (x(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) (2, 3)$x \in (0, 1) \cup (2, 3)$.

Step 2: Case II (y < 0)

If y < 0$y < 0$, the inequality reduces to (x(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) (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 ∫₀⁴ √(x) dx = 2 · (2)/(3)[x3/2]₀⁴ = (4)/(3) · 8 = (32)/(3)$= 2 \int_{0}^{4} \sqrt{x} dx = 2 \cdot \frac{2}{3}[x^{3/2}]_{0}^{4} = \frac{4}{3} \cdot 8 = \frac{32}{3}$.

Chapter Mix

Class 12 Maths: Area Under Curves Class 11 Maths: Linear Inequalities