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OpenStudy (anonymous):
intergral of sin(y) = -cos(y)
OpenStudy (anonymous):
-cos(1/2)-(-cos(0))
OpenStudy (anonymous):
-cos(1/2)+cos(0)
OpenStudy (anonymous):
with respect to what? is its respect to y then
integral of sin(y) = -cos(y)
integral of y = 1/2 y^2
integral of 0 = C
integral of 1/2 = 1/2x
OpenStudy (anonymous):
I first had to integrate arcsin(x) with respect to x from 0 to 1/2 using integration by parts....then it ask me to Calculate the area a second way by integrating with respect to y (using horizontal rectangles to
approximate area instead of vertical rectangles.)
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OpenStudy (anonymous):
was i right to assume this ment i noew integrate sin(y)=x with respect to y from 0 to 1/2?
OpenStudy (anonymous):
that should be right
OpenStudy (anonymous):
what is right?
OpenStudy (anonymous):
to do it another way yes you can use horizontal rectsangle and integrate with respect to y
and it will give you the same answer
OpenStudy (anonymous):
how do i do that?
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