integrate:sin(y),y,0,(1/2)
intergral of sin(y) = -cos(y)
-cos(1/2)-(-cos(0))
-cos(1/2)+cos(0)
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
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.)
was i right to assume this ment i noew integrate sin(y)=x with respect to y from 0 to 1/2?
that should be right
what is right?
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
how do i do that?
thank you though
Join our real-time social learning platform and learn together with your friends!