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The graph of r= 2 sin 30 is shown to the right. Write an integral which can be used to find the area of one of the "petals". Calculate and determine what that area is.

The graph of r= 2 sin 30 is shown to the right. Write an integral which can be used to find the area of one of the "petals". Calculate and determine what that area is.

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter7: Integration
Section7.3: Area And The Definite Integral
Problem 22E
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9. The graph of r = 2 sin 30 is shown to the right. Write an integral which can be used to find the area of one of the "petals". Calculate and determine what that area is.
Transcribed Image Text:9. The graph of r = 2 sin 30 is shown to the right. Write an integral which can be used to find the area of one of the "petals". Calculate and determine what that area is.
Interval of Function Convergence 1 - = 1 - (x – 1) + (x – 1)2 – (x – 1)3 + (x – 1)4 –... + (-1)"(x - 1)" + . .. 0 < x < 2 1 = 1- x + x? - x + x* - x +... + (-1)" x" +... -1 <x < 1 (x In x = (x - 1) - 1)2 (x - 1)3 (x – 1)4 (-1)"-'(x – 1)" 0 < xs 2 +... + +... 3 4 x2 et = 1 + x + 2! - 00 <x < c 3! 4! 5! n! 1)"x2n+ 1 (2n + 1)! x7 sin x = x - 3! - 00 <X < o0 5! 7! 9! (-1)" x2" (2n)! x4 cos x = 1 +... - 00 < x < 0 2! 4! 6! 8! x7 (-1)" x2n+1 + arctan x = x - 3 -1 sxs 1 +.. 7 9 2n + 1 1•3• 5x 2.4• 6• 7 1• 3x (2n)!x2n+1 (2"n!)2(2n + 1) arcsin x = x + +. . . + -1 sx s 1 2.3 2.4•5 k(k – 1)x2 k(k – 1)(k – 2)x³ k(k - 1)..· (k - n + 1)x" (1 + x)* = 1 + kx + -1 <x < 1* +... 2! 3! n! * The convergence at X = ±1 depends on the value of k.
Transcribed Image Text:Interval of Function Convergence 1 - = 1 - (x – 1) + (x – 1)2 – (x – 1)3 + (x – 1)4 –... + (-1)"(x - 1)" + . .. 0 < x < 2 1 = 1- x + x? - x + x* - x +... + (-1)" x" +... -1 <x < 1 (x In x = (x - 1) - 1)2 (x - 1)3 (x – 1)4 (-1)"-'(x – 1)" 0 < xs 2 +... + +... 3 4 x2 et = 1 + x + 2! - 00 <x < c 3! 4! 5! n! 1)"x2n+ 1 (2n + 1)! x7 sin x = x - 3! - 00 <X < o0 5! 7! 9! (-1)" x2" (2n)! x4 cos x = 1 +... - 00 < x < 0 2! 4! 6! 8! x7 (-1)" x2n+1 + arctan x = x - 3 -1 sxs 1 +.. 7 9 2n + 1 1•3• 5x 2.4• 6• 7 1• 3x (2n)!x2n+1 (2"n!)2(2n + 1) arcsin x = x + +. . . + -1 sx s 1 2.3 2.4•5 k(k – 1)x2 k(k – 1)(k – 2)x³ k(k - 1)..· (k - n + 1)x" (1 + x)* = 1 + kx + -1 <x < 1* +... 2! 3! n! * The convergence at X = ±1 depends on the value of k.
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