Find the area bounded on the left by the y-ax | vedclass

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

Question

(B) The region is bounded by $x=0$ (the $y-$axis) on the left,$x=\frac{\pi}{2}$ on the right,and the $x-$axis $(y=0)$ below.The upper boundary is defi

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(B) The region is bounded by $x=0$ (the $y-$axis) on the left,$x=\frac{\pi}{2}$ on the right,and the $x-$axis $(y=0)$ below.The upper boundary is defined by $y = \cos x$ for $0 \le x \le \frac{\pi}{4}$ and $y = \sin x$ for $\frac{\pi}{4} \le x \le \frac{\pi}{2}$,since $\cos x = \sin x$ at $x = \frac{\pi}{4}$.The required area $A$ is given by:$A = \int_{0}^{\pi/4} \cos x \, dx + \int_{\pi/4}^{\pi/2} \sin x \, dx$$A = [\sin x]_{0}^{\pi/4} + [-\cos x]_{\pi/4}^{\pi/2}$$A = (\sin(\frac{\pi}{4}) - \sin(0)) + (-(\cos(\frac{\pi}{2}) - \cos(\frac{\pi}{4})))$$A = (\frac{1}{\sqrt{2}} - 0) + (-(0 - \frac{1}{\sqrt{2}}))$$A = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$

Find the area bounded on the left by the $y-$axis,below by the $x-$axis,on the right by $x = \frac{\pi}{2}$,above on the left by $y = \cos x$,and above on the right by $y = \sin x$.

Similar Questions

The area (in sq. units) of the region enclosed between the parabola $y^{2}=2x$ and the line $x+y=4$ is

The area (in sq. units) of the region described by $\{(x, y) : y^2 \leq 2x \text{ and } y \geq 4x - 1\}$ is

The area of the region bounded by the curves $y=x^3$, $y=x^2$ and the lines $x=0$ and $x=2$ is

The area bounded by the curves $y^2 = 4x$ and $x^2 = 4y$ is

The area (in sq. units) of the region bounded by the curves $y=3x+1$,$y=4x+1$ and the line $x=3$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module