Area under Curves
The region is bounded by three primary curves in the first quadrant (x≥0$x\ge0$):
y ≥ 1$y \geq 1$ (horizontal line)
y ≤ x²$y \leq x^2$ (upward-opening parabola)
xy ≤ 8 y ≤ (8)/(x)$xy \leq 8 \implies y \leq \frac{8}{x}$ (rectangular hyperbola)
Find points of intersection:
Intersection of y = x²$y = x^2$ and y = (8)/(x)$y = \frac{8}{x}$:
x² = (8)/(x) x³ = 8 x = 2$$x^2 = \frac{8}{x} \implies x^3 = 8 \implies x = 2$$
At x = 2$x = 2$, y = 4$y = 4$. So they meet at (2, 4)$(2, 4)$.
Intersection of y = x²$y = x^2$ and y = 1$y = 1$:
x² = 1 x = 1$$x^2 = 1 \implies x = 1$$
Intersection of y = (8)/(x)$y = \frac{8}{x}$ and y = 1$y = 1$:
1 = (8)/(x) x = 8$$1 = \frac{8}{x} \implies x = 8$$
The region spans from x = 1$x = 1$ to x = 8$x = 8$. The upper boundary changes at x = 2$x = 2$.
Step 1: Set up Integrals
For x in [1, 2]$x \in [1, 2]$, the upper curve is y = x²$y = x^2$ and the lower is y = 1$y = 1$.
For x in [2, 8]$x \in [2, 8]$, the upper curve is y = (8)/(x)$y = \frac{8}{x}$ and the lower is y = 1$y = 1$.
Q5jee_main_2024_31_jan_morningArea bounded by Parabolas and Inequalities
The area of the region (x,y): y² ≤ 4x, x < 4, (xy(x - 1)(x - 2))/((x - 3)(x - 4)) > 0, x ≠ 3$\left\{(x,y): y^2 \le 4x, x < 4, \frac{xy(x - 1)(x - 2)}{(x - 3)(x - 4)} > 0, x \neq 3\right\}$ is
A.(16)/(3)$$\frac{16}{3}$$
B.(64)/(3)$$\frac{64}{3}$$
C.(8)/(3)$$\frac{8}{3}$$
D.(32)/(3)$$\frac{32}{3}$$
Solution
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.
Area bounded by Parabolas and Inequalities diagram for Q5 - JEE Main 2024 Morning
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
More Application of Integrals Questions — jee_main_2026_28_january_morning
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
