The area of the region, bounded by the curves | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The function $y = \sin^{-1} x$ is defined for $-1 \leq x \leq 1$. In the first quadrant, we have $0 \leq x \leq 1$. Since $x(1-x) \geq 0$ for $x \

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(C) The function $y = \sin^{-1} x$ is defined for $-1 \leq x \leq 1$. In the first quadrant, we have $0 \leq x \leq 1$. Since $x(1-x) \geq 0$ for $x \in [0, 1]$, the curve $y_1 = \sin^{-1} x + x(1-x)$ lies above the curve $y_2 = \sin^{-1} x - x(1-x)$. The intersection points are found by setting $y_1 = y_2$, which gives $2x(1-x) = 0$, implying $x = 0$ or $x = 1$. The required area is given by the integral: $A = \int_{0}^{1} (y_1 - y_2) dx = \int_{0}^{1} [(\sin^{-1} x + x - x^2) - (\sin^{-1} x - x + x^2)] dx$ $A = \int_{0}^{1} (2x - 2x^2) dx = 2 \int_{0}^{1} (x - x^2) dx$ $A = 2 \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1} = 2 \left( \frac{1}{2} - \frac{1}{3} \right) = 2 \left( \frac{1}{6} \right) = \frac{1}{3}$.

The area of the region, bounded by the curves $y=\sin ^{-1} x+x(1-x)$ and $y=\sin ^{-1} x-x(1-x)$ in the first quadrant, is

Similar Questions

The area (in sq. units) of the region $\{ (x,y) : x \ge 0, x + y \le 3, x^2 \le 4y \text{ and } y \le 1 + \sqrt{x} \}$ is:

Let the area enclosed between the curves $|y|=1-x^2$ and $x^2+y^2=1$ be $\alpha$. If $9\alpha=\beta\pi+\gamma$,where $\beta$ and $\gamma$ are integers,then the value of $|\beta-\gamma|$ equals

The area bounded by the curves $y=2x^2$, $y=\max \{x-[x], x+|x|\}$ and the lines $x=0, x=2$ (in sq units) is

The area of the figure bounded by the curves $y = |x - 1|$ and $y = 3 - |x|$ is ....... $sq. \text{ unit}$.

The ratio in which the $x$-axis divides the area of the region bounded by the curves $y = x^2 - 4x$ and $y = 2x - x^2$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module