Area Under the Curve
Area Under Curves
nta_abhyas_2025
Grade 12
Question:
Find the required area bounded by $y^2 = x + 2$, $y = x - 2$, and the $y$-axis where the region passes through $(-2, 0)$ and $(0, 0)$ with vertex at $(2, 2)$ marked.
Step-by-Step Solution
Key Concept: Integrate with respect to $y$ when the parabola is expressed as $x = f(y)$ to find the area between curves.
The parabola $y^2 = x + 2$ and the line $y = x - 2$ intersect where $y^2 = y + 2 - 2$, so $y^2 = y$, giving $y = 0$ or $y = 1$. At $y = 0$: $x = -2$; at $y = 1$: $x = 3$. The required area is $\int_0^2 [(4 - y^2) - (y - 2)] dy = \int_0^2 (6 - y^2 - y) dy = [6y - \frac{y^3}{3} - \frac{y^2}{2}]_0^2 = 12 - \frac{8}{3} - 2 = 10 - \frac{8}{3} = \frac{22}{3}$ sq. units.
Correct Answer: 22/3