What is the integral of (-28t+40) ?

The integral of (-28t+40) is -14t^2+40t+C

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integral of (-28t+40)

\int (- 28 t + 40) d t

- 14 t^{2} + 40 t + C

Solution steps

\int - 28 t + 40 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 28 t d t + \int 40 d t

\int 28 t d t = 14 t^{2}

\int 40 d t = 40 t

= - 14 t^{2} + 40 t

Add a constant to the solution

= - 14 t^{2} + 40 t + C