Calculus Please Check my answers

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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+3x
f(x)=x^2+2x-12 My Answer
f(x)x^2+2x-3
f(x)=x^2+4x-8

Find the range of the function f(x)=integral from 0 to x (sqrt 4-t^2) dt
(0, 4pi)
(0, 2pi) My Answer
(-4,0)
(0,4)

  • Calculus Please Check my answers -

    ?[1,x] 2t+2 dt
    = t^2+2t [1,x]
    = (x^2+2x)-(1^2+2*1)
    = x^2+2x-3

    ?[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]

    http://www.wolframalpha.com/input/?i=range+1%2F2+%E2%88%9A(4-x%5E2)+%2B+2arcsin(x%2F2)+-+1

  • Calculus Please Check my answers -

    Well that was my issue with the second one is I originally got that answer but it isn't an option.

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