What is the integral of (3t^2-4t-4) ?

The integral of (3t^2-4t-4) is t^3-2t^2-4t+C

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integral of (3t^2-4t-4)

\int (3 t^{2} - 4 t - 4) d t

t^{3} - 2 t^{2} - 4 t + C

Solution steps

\int 3 t^{2} - 4 t - 4 d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 3 t^{2} d t - \int 4 t d t - \int 4 d t

\int 3 t^{2} d t = t^{3}

\int 4 t d t = 2 t^{2}

\int 4 d t = 4 t

= t^{3} - 2 t^{2} - 4 t

Add a constant to the solution

= t^{3} - 2 t^{2} - 4 t + C

sin^{3} (x^{3} - 2 x)

\int_{0}^{3} t e^{- t} d t

\int 6 x^{2} - 5 x + 3 d x

\int (6 sin (2 x) - 4 cos (4 x)) d x

y^{2} = 4 x, 4 x - 3 y = 4

What is the integral of (3t^2-4t-4) ?
The integral of (3t^2-4t-4) is t^3-2t^2-4t+C