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find the area enclosed by y=max(sinx,-2,cosx) and x axis between x=p/4 and x=2p

find the area enclosed by y=max(sinx,-2,cosx) and x axis between x=p/4 and x=2p

Grade:12th pass

1 Answers

Arun Kumar IIT Delhi
askIITians Faculty 256 Points
9 years ago
Hello Student,
y = max(sinx, cosx, -2)
\frac{\pi }{4}\leq x\leq \frac{5\pi }{4}, y = sinx
\frac{5\pi }{4}\leq x\leq 2\pi , y = cosx
A = \int_{\pi /4}^{5\pi /4}sinxdx + \int_{5\pi /4}^{2\pi }cosxdx
A = \int_{\pi /4}^{\pi }sinxdx + |\int_{\pi }^{5\pi /4}sinxdx |+|\int_{5\pi /4}^{3\pi /2 }cosxdx|+\int_{3\pi /2}^{2\pi }cosxdxA = (-cosx)_{\pi /4}^{\pi }+|(-cosx)_{\pi }^{5\pi /4}|+|(sinx)_{5\pi /4}^{3\pi /2}|+(sinx)_{3\pi /2}^{2\pi }
A = 4 - (2)^{-1/2}
Thanks & Regards
Arun Kumar
Btech, IIT Delhi
Askiitians Faculty

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