Calculus Please Check my answers
posted by Ke$ha .
Use the graph of f(t) = 2t +2 on the interval [−1, 4] to write the function F(x), where
f(x)= integral from 1 to x f(t) dt
f(x)=x^2+2x-12 My Answer
Find the range of the function f(x)=integral from 0 to x (sqrt 4-t^2) dt
(0, 2pi) My Answer
?[1,x] 2t+2 dt
= t^2+2t [1,x]
?[0,x] ?(4-t^2) dt
= 1/2 ?(4-t^2) + 2 arcsin(t/2) [0,x]
= 1/2 ?(4-x^2) + 2arcsin(x/2) - (1/2 ?(4-0) + 2 arcsin(0))
= 1/2 ?(4-x^2) + 2arcsin(x/2) - 1
The domain is clearly [-2,2]
The range is [-?-1,?-1]
Well that was my issue with the second one is I originally got that answer but it isn't an option.