Area of the region bounded by the curves y = | vedclass

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

Question

(D) The area $A$ is given by the integral of the absolute difference between the two curves: $A = \int_{0}^{2\pi} |x - \sin x| \, dx$.Since $x \ge \si

Vedclass · Accepted Answer

$2\pi^2$

Vedclass · Answer

(D) The area $A$ is given by the integral of the absolute difference between the two curves: $A = \int_{0}^{2\pi} |x - \sin x| \, dx$.Since $x \ge \sin x$ for all $x \in [0, 2\pi]$,the absolute value can be removed: $A = \int_{0}^{2\pi} (x - \sin x) \, dx$.Evaluating the integral: $A = \left[ \frac{x^2}{2} + \cos x \right]_{0}^{2\pi}$.Substituting the limits: $A = \left( \frac{(2\pi)^2}{2} + \cos(2\pi) \right) - \left( \frac{0^2}{2} + \cos(0) \right)$.$A = (2\pi^2 + 1) - (0 + 1) = 2\pi^2$.

Area of the region bounded by the curves $y = \sin x$ and $y = x$ between the lines $x = 0$ and $x = 2\pi$ is:

Similar Questions

The area (in sq. units) of the region described by $\{(x, y): y^2 \leq 2x\}$ and $\{y \geq 4x - 1\}$ is

The area enclosed by the curves $xy + 4y = 16$ and $x + y = 6$ is equal to:

Area of the region enclosed between the curves $x = y^2 - 1$ and $x = |y| \sqrt{1 - y^2}$ is

The area bounded between the curve $x^{2}=y$ and the line $y=4x$ is

The area of the region $\{(x, y): y^2 \leq 4x, x < 4, \frac{xy(x-1)(x-2)}{(x-3)(x-4)} > 0, x \neq 3\}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module