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 bounded by the curve $y = |x^{2}-9|$ and the line $y = 3$ is

Let $A = \{(x, y) \in R^2 : y \geq 0, 2x \leq y \leq \sqrt{4-(x-1)^2}\}$ and $B = \{(x, y) \in R \times R : 0 \leq y \leq \min \{2x, \sqrt{4-(x-1)^2}\}\}$. Then the ratio of the area of $A$ to the area of $B$ is

The area (in $sq. \, units$) of the region bounded by the curves $x^{2}+2y-1=0$,$y^{2}+4x-4=0$,and $y^{2}-4x-4=0$ in the upper half plane is $....$

The area of the region (in square units) bounded by the curves $y=x^3$,$y=x$ and $-1 \leq x \leq 1$ is:

For $x \in [0, 2\pi]$,the area of the region bounded by the curves $y = x + \sin x$ and $y = x$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module