The area of the region bounded by y=cos x,x=0 | 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 $|y|$ from $x=0$ to $x=\pi$. Since $y = \cos x$ is positive in $[0, \pi/2]$ and negative in $[\pi/2, \pi]

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) The area $A$ is given by the integral of $|y|$ from $x=0$ to $x=\pi$. Since $y = \cos x$ is positive in $[0, \pi/2]$ and negative in $[\pi/2, \pi]$,the area is: $A = \int_0^{\pi/2} \cos x \, dx + \left| \int_{\pi/2}^{\pi} \cos x \, dx ight|$ $A = [\sin x]_0^{\pi/2} + |[\sin x]_{\pi/2}^{\pi}|$ $A = (\sin(\pi/2) - \sin(0)) + |\sin(\pi) - \sin(\pi/2)|$ $A = (1 - 0) + |0 - 1|$ $A = 1 + |-1| = 1 + 1 = 2 	ext{ sq. units.}$

The area of the region bounded by $y=\cos x$,$x=0$,$x=\pi$,and the $x$-axis is ... sq. units.

Similar Questions

If the area bounded by the parabola $y^2=16ax$ and the line $y=4mx$ is $\frac{a^2}{12}$ sq. units,then the value of '$m$' is

The area of the region bounded by the $y$-axis,$y=\cos x$,and $y=\sin x$,when $0 \leq x \leq \frac{\pi}{4}$,is

Find the area of the region $\{(x, y): 0 \leq y \leq x^{2}+1, 0 \leq y \leq x+1, 0 \leq x \leq 2\}$.

The smaller area enclosed by the circle $x^{2}+y^{2}=4$ and the line $x+y=2$ is

The area (in $sq. \text{ units}$) bounded by the curve $y=x|x|$,$X$-axis and the lines $x=-1$ and $x=1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module