Math  /  Calculus

QuestionQuestion Compl
Use graphs to evaluate 02πsinxdx\int_{0}^{2 \pi} \sin x d x and 02πcosxdx\int_{0}^{2 \pi} \cos x d x. 02πsinxdx=\int_{0}^{2 \pi} \sin x d x= \square 02πcosxdx=\int_{0}^{2 \pi} \cos x d x= \square

Studdy Solution
The area under the curve of cosx\cos x from 00 to π\pi is positive and equal in magnitude to the area from π\pi to 2π2\pi, which is negative. Therefore, the integral over the full interval [0,2π][0, 2\pi] is zero.
02πcosxdx=0\int_{0}^{2 \pi} \cos x \, dx = 0
The evaluated integrals are:
02πsinxdx=0\int_{0}^{2 \pi} \sin x \, dx = \boxed{0}
02πcosxdx=0\int_{0}^{2 \pi} \cos x \, dx = \boxed{0}

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