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?

Question

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?

irishboy123 · Answer

you should get $\large [-cos y]_{\pi/4}^{3\pi/4}$

irishboy123 · Answer

so i assume algebraic error (minus signs) as you seem to have got that.....

$-(-\frac{1}{\sqrt{2 } } )-(-\frac{1}{\sqrt{2 } })$

amistre64 · Answer

-cos(3pi/4) +cos(pi/4)
 --1/sqrt2 +  1/sqrt2   =  2/sqrt2 = sqrt2

another way to see it is:  1/sqrt2 = sqrt2/2

amistre64 · Answer

yeah, he is missing the minus sign to me as well