What is the integral of (3-0.01x+0.00006x^2) ?

The integral of (3-0.01x+0.00006x^2) is 3x-0.005x^2+0.00002x^3+C

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integral of (3-0.01x+0.00006x^2)

\int (3 - 0.01 x + 0.00006 x^{2}) d x

3 x - 0.005 x^{2} + 0.00002 x^{3} + C

Solution steps

\int 3 - 0.01 x + 0.00006 x^{2} 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 3 d x - \int 0.01 x d x + \int 0.00006 x^{2} d x

\int 3 d x = 3 x

\int 0.01 x d x = 0.005 x^{2}

\int 0.00006 x^{2} d x = 0.00002 x^{3}

= 3 x - 0.005 x^{2} + 0.00002 x^{3}

Add a constant to the solution

= 3 x - 0.005 x^{2} + 0.00002 x^{3} + C

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What is the integral of (3-0.01x+0.00006x^2) ?
The integral of (3-0.01x+0.00006x^2) is 3x-0.005x^2+0.00002x^3+C