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

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(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

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$1$

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(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 ......... $sq. \text{ } unit$.

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