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anonymous
 one year ago
Find the area between a curve. X=0 ,x=siny, y=pi/4, y=3pi/4
Anyways I set it up and got cosy as my antiderivative. And when I plug in the boundaries I get sqrt2/2 both times to equal zero. My book says the answer is sqrt2 what am I doing wronf?
anonymous
 one year ago
Find the area between a curve. X=0 ,x=siny, y=pi/4, y=3pi/4 Anyways I set it up and got cosy as my antiderivative. And when I plug in the boundaries I get sqrt2/2 both times to equal zero. My book says the answer is sqrt2 what am I doing wronf?

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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1you should get \(\large [cos y]_{\pi/4}^{3\pi/4}\)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1so i assume algebraic error (minus signs) as you seem to have got that..... \((\frac{1}{\sqrt{2 } } )(\frac{1}{\sqrt{2 } })\)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1cos(3pi/4) +cos(pi/4) 1/sqrt2 + 1/sqrt2 = 2/sqrt2 = sqrt2 another way to see it is: 1/sqrt2 = sqrt2/2

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1yeah, he is missing the minus sign to me as well
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