Area Under the Curve
Area Under Curves
nta_abhyas_2025
Grade 12
Question:
The area between the curve $y = 2x^2 - s^2$, the $x$-axis and the ordinates of the two minima of the curve is
$\frac{25}{12}$ sq. units
$\frac{20}{3}$ sq. units
$\frac{119}{120}$ sq. units
$\frac{25}{9}$ sq. units
Step-by-Step Solution
Key Concept: Use symmetry properties and critical points to identify the bounds and evaluate the definite integral for area enclosed by a curve and the $x$-axis.
The curve is symmetric about the $y$-axis and touches the $x$-axis at $x = 0, \pm\frac{1}{\sqrt{2}}$. Setting $\frac{dy}{dx} = 8x^3 - 2x = 2x(4x^2 - 1) = 0$ gives critical points at $x = 0, \pm\frac{1}{2}$. Points of local minima occur at $x = \pm\frac{1}{2}$. The total area is $2\int_0^{1/\sqrt{2}} (2x^4 - x^2)\,dx = 2\left[\frac{2x^5}{5} - \frac{x^3}{3}\right]_0^{1/\sqrt{2}} = \frac{12}{35}$ sq unit from the solution shown.
Correct Answer: 12