The area of the figure enclosed by y = cos^{- | vedclass

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

Question

(A) The function $y = \cos^{-1}(\cos x)$ for $x \in [2\pi, 4\pi]$ is defined as:$y = \begin{cases} x - 2\pi, & x \in [2\pi, 3\pi] \ 4\pi - x, & x \in

Vedclass · Accepted Answer

$\frac{3}{4}\pi^2$

Vedclass · Answer

(A) The function $y = \cos^{-1}(\cos x)$ for $x \in [2\pi, 4\pi]$ is defined as:$y = \begin{cases} x - 2\pi, & x \in [2\pi, 3\pi] \ 4\pi - x, & x \in [3\pi, 4\pi] \end{cases}$The function $y = 	an^{-1} x + 	an^{-1} \frac{1}{x}$ for $x > 0$ is $y = \frac{\pi}{2}$.The area is bounded by the $x$-axis $(y=0)$, the line $y = \frac{\pi}{2}$, and the curve $y = \cos^{-1}(\cos x)$.Looking at the graph, the region is a trapezoid with parallel sides at $y = \frac{\pi}{2}$.The intersection points of $y = \cos^{-1}(\cos x)$ and $y = \frac{\pi}{2}$ are:$x - 2\pi = \frac{\pi}{2} \implies x = \frac{5\pi}{2}$$4\pi - x = \frac{\pi}{2} \implies x = \frac{7\pi}{2}$The area of the trapezoid is given by $\frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes 	ext{height}$.The parallel sides are the segment on the $x$-axis from $2\pi$ to $4\pi$ (length $2\pi$) and the segment at $y = \frac{\pi}{2}$ from $\frac{5\pi}{2}$ to $\frac{7\pi}{2}$ (length $\pi$).Area $= \frac{1}{2} 	imes (2\pi + \pi) 	imes \frac{\pi}{2} = \frac{1}{2} 	imes 3\pi 	imes \frac{\pi}{2} = \frac{3\pi^2}{4}$.

The area of the figure enclosed by $y = \cos^{-1}(\cos x)$, $x \in [2\pi, 4\pi]$, the $x$-axis, and $y = \tan^{-1} x + \tan^{-1} \frac{1}{x}$ is

Similar Questions

The area enclosed by the parabolas $y = x^2 - 1$ and $y = 1 - x^2$ is (in $/3$)

Let $y=p(x)$ be the parabola passing through the points $(-1,0), (0,1)$ and $(1,0)$. If the area of the region $\{(x, y) : (x+1)^2+(y-1)^2 \leq 1, y \leq p(x)\}$ is $A$,then $12(\pi-4A)$ is equal to $.........$.

The area (in sq. units) of the region enclosed between the parabola $y^{2}=2x$ and the line $x+y=4$ is

The area of the region bounded by the curves $y^{2}=8x$ and $y=x$ is

The area bounded by the curves $y = |x| - 1$ and $y = -|x| + 1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module