The area of the region bounded by the y-axis, | vedclass

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

Question

(C) The region is bounded by the $y$-axis $(x=0)$,$y = \cos x$,and $y = \sin x$. These curves intersect when $\cos x = \sin x$,which occurs at $x = \f

Vedclass · Accepted Answer

$ \sqrt{2} - 1 $ Sq.units

Vedclass · Answer

(C) The region is bounded by the $y$-axis $(x=0)$,$y = \cos x$,and $y = \sin x$. These curves intersect when $\cos x = \sin x$,which occurs at $x = \frac{\pi}{4}$ in the interval $[0, \frac{\pi}{2}]$.In the interval $[0, \frac{\pi}{4}]$,$\cos x \geq \sin x$.Therefore,the required area is given by:$	ext{Area} = \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) dx$$= [\sin x - (-\cos x)]_{0}^{\frac{\pi}{4}}$$= [\sin x + \cos x]_{0}^{\frac{\pi}{4}}$$= (\sin \frac{\pi}{4} + \cos \frac{\pi}{4}) - (\sin 0 + \cos 0)$$= (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1)$$= \frac{2}{\sqrt{2}} - 1$$= \sqrt{2} - 1 	ext{ sq. units}$

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

Similar Questions

The area bounded by the parabola $y^2 = 4ax$,its axis and two ordinates $x = 4$ and $x = 9$ is

Area enclosed by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is . . . . . . .

The area of the region bounded by the curve $y = \tan x$,the tangent drawn to the curve at $x = \frac{\pi}{4}$,and the $x$-axis is:

The area bounded by the parabola $y = 4x^2$,the $y$-axis,and the lines $y = 1$ and $y = 4$ is:

The area of the region (in square units) bounded by the curve ${x^2} = 4y$,the line $x = 2$,and the $x$-axis is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module