What is the integral of-0.006x^2+0.4x ?

The integral of-0.006x^2+0.4x is -0.002x^3+0.2x^2+C

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integral of-0.006x^2+0.4x

\int - 0.006 x^{2} + 0.4 x d x

- 0.002 x^{3} + 0.2 x^{2} + C

Solution steps

\int - 0.006 x^{2} + 0.4 x d x

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 x^{2} d x + \int 0.4 x d x

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

\int 0.4 x d x = 0.2 x^{2}

= - 0.002 x^{3} + 0.2 x^{2}

Add a constant to the solution

= - 0.002 x^{3} + 0.2 x^{2} + C

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What is the integral of-0.006x^2+0.4x ?
The integral of-0.006x^2+0.4x is -0.002x^3+0.2x^2+C