0

Guest

Courses New

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
8 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

Think You Can Provide A Better Answer ?

Other Related Questions on Integral Calculus

View all Questions »

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More