(a) Sketch y=\sin x for x = −π to x = π.

(b) Calculate \int_{-\pi}^\pi \sin x d x and comment on your findings.

(c) Calculate the area enclosed by y=\sin x and the x axis between x = −π and x = π.

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(a) A graph of y = sin x between x = −π and x = π is shown in Figure 13.12.

(b) \int_{-\pi}^\pi \sin x d x=[-\cos x]_{-\pi}^\pi=-\cos (\pi)+\cos (-\pi)=0

Examining Figure 13.12 we see that the positive and negative contributions have cancelled each other out; that is, the area above the x axis is equal in size to the area below the x axis.

(c) From (b) the area above the x axis is equal in size to the area below the x axis. From Example 13.7(c) the area above the x axis is 2. Hence the total area enclosed by y=\sin x and the x axis from x = −π to x = π is 4.

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