What is the integral of 1.8x-0.12x^2 ?

The integral of 1.8x-0.12x^2 is 0.9x^2-0.04x^3+C

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integral of 1.8x-0.12x^2

\int 1.8 x - 0.12 x^{2} d x

0.9 x^{2} - 0.04 x^{3} + C

Solution steps

\int 1.8 x - 0.12 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 1.8 x d x - \int 0.12 x^{2} d x

\int 1.8 x d x = 0.9 x^{2}

\int 0.12 x^{2} d x = 0.04 x^{3}

= 0.9 x^{2} - 0.04 x^{3}

Add a constant to the solution

= 0.9 x^{2} - 0.04 x^{3} + C

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What is the integral of 1.8x-0.12x^2 ?
The integral of 1.8x-0.12x^2 is 0.9x^2-0.04x^3+C