Mathematics, 20.10.2019 19:50 priceisright11401
Find the area of the region bounded by the curves y= sin^-1(x/6), y=0, and x=6 obtained by integrating with respect to y. your work must include the definite integral and the antiderivative.
Answers: 3
![\bf -------------------------------\\\\ \displaystyle \int\limits_{0}^{\frac{\pi }{2}}\ [6-6sin(y)]dy\implies \left. \cfrac{}{} 3y^2+6cos(y) \right]_{0}^{\frac{\pi }{2}} \\\\\\ \left[ \cfrac{3\pi 2}{4}+0 \right]-[0+6]\implies \cfrac{3\pi^2}{4}-6\approx 1.4022033](/tpl/images/0337/8410/9bf04.png)
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Find the area of the region bounded by the curves y= sin^-1(x/6), y=0, and x=6 obtained by integrati...
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