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

The value of $a$ $(a > 0)$ for which the area bounded by the curves $y = \frac{x}{6} + \frac{1}{x^2}$,$y = 0$,$x = a$,and $x = 2a$ has the least value,is

The area of the region bounded by the curve $y=2x-x^2$ and the $x$-axis is

The area of the region bounded by the curve $x+2y=8$,the $X$-axis,and the lines $x=1$ and $x=5$ using integration is . . . . . . sq. units.

Area of the region bounded by the ellipse $9x^2 + 16y^2 = 1$ in the first quadrant is . . . . . . sq. units.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module