+91-87964 74404
FlagBACK
Question icon
Grade 12th passIntegral Calculus

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

Profile image of John Doe
12 Years agoGrade 12th pass
Answers icon

1 Answer

Profile image of Arun Kumar
12 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

Other Related Questions on integral calculus

question mark
The area of a square field is 640000cm2.what is its area in hectares
integral calculus
1 Answer Available

Last Activity: 3 Years ago

question mark
Please solve this integration. I am facing difficulty in it
 How to solve this. Please solve anyone. Humble request
integral calculus
1 Answer Available

Last Activity: 4 Years ago

question mark
Using laplace transform, solve d2y/dt2 + dy/dt = t2 +2t given that y=4 and y’=-2 and t=0
integral calculus
1 Answer Available

Last Activity: 4 Years ago

question mark
If 4th term of an A.p is zero, show that the 8th term is double the 6th term
 
integral calculus
1 Answer Available

Last Activity: 4 Years ago