Use the graph of f(t) = 2t + 3 on the interval [-3, 6] to write the function F(x), where F(x)= ∫f(t) dt where a=3 b=x.

F(x) = 2x^2 + 6x
F(x) = 2x + 3
F(x) = x^2 + 3x + 54
F(x) = x^2 + 3x - 18
Honestly have no idea where to start. Do i take the derivative of that or what?

asked by Henry
  1. surely you can integrate

    ∫2t+3 dt

    The tricky part is what to do with the limits. That means you want

    F(x) = ∫[3,x] 2t+3 dt

    Now just evaluate normally, but using x as the upper limit, instead of a number. Then you have

    F(x) = t^2+3t [3,x]
    = (x^2+3x)-(3^2+3*3)
    = x^2+3x-18

    posted by Steve

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