The area under the curve y = |cos x - sin x|, | vedclass

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

Question

(B) The function is $y = |\cos x - \sin x|$.In the interval $[0, \pi/4]$,$\cos x \geq \sin x$,so $y = \cos x - \sin x$.In the interval $[\pi/4, \pi/2]

Vedclass · Accepted Answer

$2\sqrt{2} - 2$

Vedclass · Answer

(B) The function is $y = |\cos x - \sin x|$.In the interval $[0, \pi/4]$,$\cos x \geq \sin x$,so $y = \cos x - \sin x$.In the interval $[\pi/4, \pi/2]$,$\sin x \geq \cos x$,so $y = \sin x - \cos x$.The required area is given by:$A = \int_{0}^{\pi/4} (\cos x - \sin x) \, dx + \int_{\pi/4}^{\pi/2} (\sin x - \cos x) \, dx$$A = [\sin x + \cos x]_{0}^{\pi/4} + [-\cos x - \sin x]_{\pi/4}^{\pi/2}$$A = ((\sin(\pi/4) + \cos(\pi/4)) - (\sin(0) + \cos(0))) + ((-\cos(\pi/2) - \sin(\pi/2)) - (-\cos(\pi/4) - \sin(\pi/4)))$$A = ((\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1)) + ((0 - 1) - (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}))$$A = (\sqrt{2} - 1) + (-1 + \sqrt{2})$$A = 2\sqrt{2} - 2$ square units.

The area under the curve $y = |\cos x - \sin x|$,$0 \leq x \leq \frac{\pi}{2}$,and above the $x$-axis is

Similar Questions

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

Area enclosed by the ellipse $9x^2 + 4y^2 = 1$ in the first quadrant is . . . . . . .

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

The area enclosed between the curves $y=x|x|$ and $y=x-|x|$ is :

The area (in sq. units) of the region bounded by the curve $y = 2x - x^2$ and the $X$-axis is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module