What is the integral of-0.92t+1.86 ?

The integral of-0.92t+1.86 is -0.46t^2+1.86t+C

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integral of-0.92t+1.86

\int - 0.92 t + 1.86 d t

- 0.46 t^{2} + 1.86 t + C

Solution steps

\int - 0.92 t + 1.86 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 0.92 t d t + \int 1.86 d t

\int 0.92 t d t = 0.46 t^{2}

\int 1.86 d t = 1.86 t

= - 0.46 t^{2} + 1.86 t

Add a constant to the solution

= - 0.46 t^{2} + 1.86 t + C

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What is the integral of-0.92t+1.86 ?
The integral of-0.92t+1.86 is -0.46t^2+1.86t+C