√ - is my square root symbol.
For an answer i got 2pi^2
Is that a good answer?
just doesn't look right to me...
Any one can help? thank you
sqrt[ sin^2(theta) + cos^2(theta) ]
= sqrt(1)
= 1.
2pi int(0, pi) [ 1 + cos(theta) ] d theta
= [ theta + sin(theta) ](0, pi)
= 2pi^2.
It's a good answer if the question is right, but you don't often get anything like sqrt[ sin^2(theta) + cos^2(theta) ] in integration questions.
Copyright © 2024 1QUIZZ.COM - All rights reserved.
Answers & Comments
Verified answer
sqrt[ sin^2(theta) + cos^2(theta) ]
= sqrt(1)
= 1.
2pi int(0, pi) [ 1 + cos(theta) ] d theta
= [ theta + sin(theta) ](0, pi)
= 2pi^2.
It's a good answer if the question is right, but you don't often get anything like sqrt[ sin^2(theta) + cos^2(theta) ] in integration questions.