The area of the region bounded by the curve y | vedclass

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

Question

(C) The area $A$ is given by the integral of the absolute value of the function over the interval $[0, 2]$.$A = \int_{0}^{2} |\sin(\pi x)| \, dx$Since

Vedclass · Accepted Answer

$\frac{4}{\pi}$

Vedclass · Answer

(C) The area $A$ is given by the integral of the absolute value of the function over the interval $[0, 2]$.$A = \int_{0}^{2} |\sin(\pi x)| \, dx$Since $\sin(\pi x) \ge 0$ for $x \in [0, 1]$ and $\sin(\pi x) \le 0$ for $x \in [1, 2]$,we split the integral:$A = \int_{0}^{1} \sin(\pi x) \, dx + \int_{1}^{2} -\sin(\pi x) \, dx$Evaluating the first part:$\int_{0}^{1} \sin(\pi x) \, dx = [-\frac{\cos(\pi x)}{\pi}]_{0}^{1} = -\frac{1}{\pi}(\cos(\pi) - \cos(0)) = -\frac{1}{\pi}(-1 - 1) = \frac{2}{\pi}$Evaluating the second part:$\int_{1}^{2} -\sin(\pi x) \, dx = [\frac{\cos(\pi x)}{\pi}]_{1}^{2} = \frac{1}{\pi}(\cos(2\pi) - \cos(\pi)) = \frac{1}{\pi}(1 - (-1)) = \frac{2}{\pi}$Total Area $A = \frac{2}{\pi} + \frac{2}{\pi} = \frac{4}{\pi}$ sq. units.Therefore,the correct option is $C$.

The area of the region bounded by the curve $y = \sin(\pi x)$ and the $X$-axis for $x \in [0, 2]$ is . . . . . . sq. units.

Similar Questions

The line $y = mx$ bisects the area enclosed by the curve $y = 1 + 4x - x^2$ and the lines $x = 0, x = \frac{3}{2}$ and $y = 0$. Then the value of $m$ is:

The area bounded by the parabola $y=x^2$ and the line $y=x$ is

The area of the region bounded by the curve $y = \tan x$,the tangent drawn to the curve at $x = \frac{\pi}{4}$,and the $x$-axis is:

If the straight line $x=b$ divides the area enclosed by $y=(1-x)^2$,$y=0$,and $x=0$ into two parts $R_1 (0 \leq x \leq b)$ and $R_2 (b \leq x \leq 1)$ such that $R_1 - R_2 = \frac{1}{4}$,then $b$ equals:

The area of the region bounded by the curves $y = \sin x$,$y = \cos x$ and $x = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module