int_0^{2pi} (sin x + cos x) , dx = | vedclass

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(A) We need to evaluate the definite integral $I = \int_0^{2\pi} (\sin x + \cos x) \, dx$. Applying the fundamental theorem of calculus,we find the an

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$0$

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(A) We need to evaluate the definite integral $I = \int_0^{2\pi} (\sin x + \cos x) \, dx$. Applying the fundamental theorem of calculus,we find the antiderivative of $(\sin x + \cos x)$,which is $(-\cos x + \sin x)$. Evaluating this at the limits $0$ and $2\pi$: $I = [-\cos x + \sin x]_0^{2\pi}$ $I = (-\cos(2\pi) + \sin(2\pi)) - (-\cos(0) + \sin(0))$ Since $\cos(2\pi) = 1$,$\sin(2\pi) = 0$,$\cos(0) = 1$,and $\sin(0) = 0$: $I = (-1 + 0) - (-1 + 0)$ $I = -1 + 1 = 0$ Thus,the correct option is $A$.

$\int_0^{2\pi} (\sin x + \cos x) \, dx = $

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