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

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

Vedclass · Accepted Answer

$2$

Vedclass · Answer

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

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

Similar Questions

The area bounded by the $y-$axis,$y=\cos x$ and $y=\sin x$ when $0 \leq x \leq \frac{\pi}{2}$ is

The area between $x=y^{2}$ and $x=4$ is divided into two equal parts by the line $x=a$. Find the value of $a$.

The area bounded by the curve $y = \sin x$ between $x = -\pi/2$ and $x = \pi/2$ is . . . . . . .

Find the area under the given curve $y=x^{2}$ bounded by the lines $x=1$,$x=2$ and the $x$-axis.

The area (in sq units) of the region bounded by $x=-1$,$x=2$,$y=x^2+1$ and $y=2x-2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module