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nata0808 [166]
3 years ago
12

The area of the region in the first quadrant between the graph of y = k*sin(x) and x-axis (where k > 0 ) on the closed interv

al [0, \pi] is?
Mathematics
1 answer:
astraxan [27]3 years ago
8 0
Note that 0\le k\sin x\le k in the given closed interval, so the area is exactly given by the definite integral

\displaystyle\int_0^\pi k\sin x\,\mathrm dx

(no absolute values needed!)

Integrating gives

-k\cos x\bigg|_{x=0}^{x=\pi}=-k(\cos\pi-\cos0)=-k(-1-1)=2k
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