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

Question

Grade 12th passIntegral Calculus

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

John Doe

12 Years agoGrade 12th pass

Answer 1

Hello Student,

$y = max(sinx, cosx, -2)$

$\frac{\pi }{4}\leq x\leq \frac{5\pi }{4}, y = sinx$

$\frac{5\pi }{4}\leq x\leq 2\pi , y = cosx$

$A = \int_{\pi /4}^{5\pi /4}sinxdx + \int_{5\pi /4}^{2\pi }cosxdx$

$A = \int_{\pi /4}^{\pi }sinxdx + |\int_{\pi }^{5\pi /4}sinxdx |+|\int_{5\pi /4}^{3\pi /2 }cosxdx|+\int_{3\pi /2}^{2\pi }cosxdx$ $A = (-cosx)_{\pi /4}^{\pi }+|(-cosx)_{\pi }^{5\pi /4}|+|(sinx)_{5\pi /4}^{3\pi /2}|+(sinx)_{3\pi /2}^{2\pi }$

$A = 4 - (2)^{-1/2}$

Thanks & Regards
Arun Kumar
Btech, IIT Delhi
Askiitians Faculty

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

John Doe

1 Answer

Arun Kumar

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