What is the integral of-0.006t^2+0.2t ?

The integral of-0.006t^2+0.2t is -0.002t^3+0.1t^2+C

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integral of-0.006t^2+0.2t

\int - 0.006 t^{2} + 0.2 t d t

- 0.002 t^{3} + 0.1 t^{2} + C

Solution steps

\int - 0.006 t^{2} + 0.2 t 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 0.006 t^{2} d t + \int 0.2 t d t

\int 0.006 t^{2} d t = 0.002 t^{3}

\int 0.2 t d t = 0.1 t^{2}

= - 0.002 t^{3} + 0.1 t^{2}

Add a constant to the solution

= - 0.002 t^{3} + 0.1 t^{2} + C

\frac{\partial}{\partial x} (3 x - y - 1)

\int \frac{1}{( x ^{2} + 5 )} d x

\int \frac{y}{x ^{2} y ^{2} + 1} d x

9 - x^{2}

3 x - 5

What is the integral of-0.006t^2+0.2t ?
The integral of-0.006t^2+0.2t is -0.002t^3+0.1t^2+C