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: $A = \int_{0}^{3\pi/2} |\cos x| \, dx$. Since $\cos x$ is positive in

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) The area $A$ is given by the integral of the absolute value of the function: $A = \int_{0}^{3\pi/2} |\cos x| \, dx$. Since $\cos x$ is positive in $[0, \pi/2]$ and negative in $[\pi/2, 3\pi/2]$,we split the integral: $A = \int_{0}^{\pi/2} \cos x \, dx + \int_{\pi/2}^{3\pi/2} -\cos x \, dx$. Evaluating the first part: $[\sin x]_{0}^{\pi/2} = \sin(\pi/2) - \sin(0) = 1 - 0 = 1$. Evaluating the second part: $-[\sin x]_{\pi/2}^{3\pi/2} = -(\sin(3\pi/2) - \sin(\pi/2)) = -(-1 - 1) = -(-2) = 2$. Total area $A = 1 + 2 = 3$ square units.

The area of the region bounded by the curve $y = \cos x$,$x = 0$,and $x = \frac{3\pi}{2}$ is:

Similar Questions

Area enclosed by the ellipse $9x^2 + 4y^2 = 36$ is . . . . . . sq. units. (in $\pi$)

Find the area bounded by the curves $(x-1)^{2}+y^{2}=1$ and $x^{2}+y^{2}=1$.

The area of the region bounded by the parabola $x^2 = 12y$ and its latus rectum is . . . . . . sq. units.

Using the method of integration,find the area bounded by the curve $|x|+|y|=1$.

The area of the smaller region bounded by the circle $x^2 + y^2 = 4$ and the line $x = 1$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module