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Find the most general antiderivative of the function. (check your answer by differentiation. use c for the constant of the antiderivative. ) g(t) = 8 t t2 t

Sagot :

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).

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