What is the integral of 0.0003x^2-0.12x+20 ?

The integral of 0.0003x^2-0.12x+20 is 0.0001x^3-0.06x^2+20x+C

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integral of 0.0003x^2-0.12x+20

\int 0.0003 x^{2} - 0.12 x + 20 d x

0.0001 x^{3} - 0.06 x^{2} + 20 x + C

Solution steps

\int 0.0003 x^{2} - 0.12 x + 20 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 0.0003 x^{2} d x - \int 0.12 x d x + \int 20 d x

\int 0.0003 x^{2} d x = 0.0001 x^{3}

\int 0.12 x d x = 0.06 x^{2}

\int 20 d x = 20 x

= 0.0001 x^{3} - 0.06 x^{2} + 20 x

Add a constant to the solution

= 0.0001 x^{3} - 0.06 x^{2} + 20 x + C

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What is the integral of 0.0003x^2-0.12x+20 ?
The integral of 0.0003x^2-0.12x+20 is 0.0001x^3-0.06x^2+20x+C