Asked by Slaya
Find the area of the region bounded by the curves y = sin x, y = csc^2x, x = pi/4, and x = (3pi)/4.
Answers
Answered by
Reiny
A quick peek at our problem:
https://www.wolframalpha.com/input/?i=plot+y+%3D+sin+x%2C+y+%3D+csc%5E2x
looks like they intersect at x = π/2 and there is symmetry
so we can take
area = 2∫ (csc^2 x - sinx) dx from π/4 to π/2
recalling that the derivative of cotx = -sec^2 x
= 2[ (- cotx + cosx) ] from π/4 to π/2
= 2 ( (-cot π/2 + cos π/2 - (-cot π/4 + cos π/4) )
= 2( 0 + 0 - (-1 + √2/2) )
= 2(1 - √2/2)
= 2 - √2
https://www.wolframalpha.com/input/?i=plot+y+%3D+sin+x%2C+y+%3D+csc%5E2x
looks like they intersect at x = π/2 and there is symmetry
so we can take
area = 2∫ (csc^2 x - sinx) dx from π/4 to π/2
recalling that the derivative of cotx = -sec^2 x
= 2[ (- cotx + cosx) ] from π/4 to π/2
= 2 ( (-cot π/2 + cos π/2 - (-cot π/4 + cos π/4) )
= 2( 0 + 0 - (-1 + √2/2) )
= 2(1 - √2/2)
= 2 - √2
Answered by
Slaya
Thank you!
Answered by
oobleck
This is so easy how do u not know
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