The most general antiderivative of the given function g(t) is (8t + t³/3 + t²/2 + c).
The antiderivative of a function is the inverse function of a derivative.
This inverse function of the derivative is called integration.
Here the given function is: g(t) = 8 + t² + t
Therefore, the antiderivative of the given function is
∫g(t) dt
= ∫(8 + t² + t) dt
= ∫8 dt + ∫t² dt + ∫t dt
= [8t⁽⁰⁺¹⁾/(0+1) + t⁽²⁺¹⁾/(2+1) + t⁽¹⁺¹⁾/(1+1) + c]
= (8t + t³/3 + t²/2 + c)
Here 'c' is the constant.
Again, differentiating the result, we get:
d/dt(8t + t³/3 + t²/2 + c)
= [8 ˣ 1 ˣ t⁽¹⁻¹⁾ + 3 ˣ t⁽³⁻¹⁾/3 + 2 ˣ t⁽²⁻¹⁾/2 + 0]
= 8 + t² + t
= g(t)
The antiderivative of the given function g(t)is (8t + t³/3 + t²/2 + c).
Learn more about antiderivative here: https://brainly.com/question/20565614
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