Area under the curve y = sin 2x + cos 2x betw | vedclass

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

Question

(B) The required area is given by the definite integral of the function $y = \sin 2x + \cos 2x$ from $x = 0$ to $x = \frac{\pi}{4}$.$	ext{Area} = \in

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) The required area is given by the definite integral of the function $y = \sin 2x + \cos 2x$ from $x = 0$ to $x = \frac{\pi}{4}$.$	ext{Area} = \int_{0}^{\frac{\pi}{4}} (\sin 2x + \cos 2x) \, dx$Integrating the terms,we get:$= \left[ -\frac{\cos 2x}{2} + \frac{\sin 2x}{2} ight]_{0}^{\frac{\pi}{4}}$Applying the limits:$= \frac{1}{2} \left[ (-\cos(\frac{\pi}{2}) + \sin(\frac{\pi}{2})) - (-\cos(0) + \sin(0)) ight]$Since $\cos(\frac{\pi}{2}) = 0$,$\sin(\frac{\pi}{2}) = 1$,$\cos(0) = 1$,and $\sin(0) = 0$:$= \frac{1}{2} [ (0 + 1) - (-1 + 0) ]$$= \frac{1}{2} [ 1 + 1 ]$$= \frac{1}{2} 	imes 2 = 1 \, sq. 	ext{ } unit$.

Area under the curve $y = \sin 2x + \cos 2x$ between $x = 0$ and $x = \frac{\pi}{4}$ is ......... $sq. \text{ } unit$.

Similar Questions

If $y = 2 \sin x + \sin 2x$ for $0 \le x \le 2 \pi ,$ then the area enclosed by the curve and the $x-$axis is

The area enclosed by the graph of the function $y = \ln^2 x - 1$ lying in the $4^{th}$ quadrant is:

The area lying between the curves $y^{2}=4x$ and $y=2x$ is

The area bounded by the curve $y=\sin \left(\frac{x}{3}\right)$,the $x$-axis,and the lines $x=0$ and $x=3 \pi$ is

The area bounded by the curve $y = x$,the $x$-axis,and the ordinates $x = -1$ to $x = 2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module