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the fundamental theorem of calculus,

f(x)=∫(0,x) t^3+2t^2+ 2 dt,
and find f"(x).

my answer was:

it said its wrong. i don't know how solve with interval, ∫(0,x)


    I must admit I have never seen that notation before

    is this ∫ (t^3 + 2t^2 + 2) dt from 0 to x ??
    if so,

    = (1/4)t^4 + (2/3)t^3 + 2t | from 0 to x

    = (1/4)x^4 + (2/3)x^3 + 2x

    so f(x) = (1/4)x^4 + (2/3)x^3 + 2x
    f'(x) = x^3 + 2x^2 + 2 ---- as expected, the original
    f''(x) = 3x^2 + 4x


    yea it is ∫ (t^3 + 2t^2 + 2) dt from 0 to x. thanks for ur help

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