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

(B) The area $A$ is given by the integral of the absolute value of the function over the interval $[\frac{\pi}{2}, \frac{3\pi}{2}]$.$A = \int_{\frac{\

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The area $A$ is given by the integral of the absolute value of the function over the interval $[\frac{\pi}{2}, \frac{3\pi}{2}]$.$A = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} |\cos x| \, dx$In the interval $[\frac{\pi}{2}, \frac{3\pi}{2}]$,$\cos x$ is negative in the interval $(\frac{\pi}{2}, \frac{3\pi}{2}]$.Thus,$|\cos x| = -\cos x$.$A = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} -\cos x \, dx$$A = -[\sin x]_{\frac{\pi}{2}}^{\frac{3\pi}{2}}$$A = -[\sin(\frac{3\pi}{2}) - \sin(\frac{\pi}{2})]$$A = -[-1 - 1]$$A = -[-2] = 2$ sq. units.Therefore,the correct option is $B$.

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

Similar Questions

Area of the region bounded by the curve $y^2 = 4x$,$Y$-axis and the line $y = 3$ is . . . . . . sq. units.

The area of the region bounded by the curve $y=2x-x^2$ and the line $y=x$ is . . . . . . square units.

The line $x=\frac{\pi}{4}$ divides the area of the region bounded by $y=\sin x$, $y=\cos x$ and the $x$-axis $\left(0 \leq x \leq \frac{\pi}{2}\right)$ into two regions of areas $A_1$ and $A_2$. Then $A_1 : A_2$ equals (in $: 1$)

The area bounded by the curve $x(x^2 + p) = y - 1$ and the line $y = 1$ is:

The area of the region bounded by the curves $y = |x - 4|$,$x = 3$,$x = 5$,and the $X$-axis is

Vedclass Test Series

Exam Paper Generator

Online Exam Module