What is the integral of 0.024(0.06x^2-4x)+150 ?

The integral of 0.024(0.06x^2-4x)+150 is 0.024(0.02x^3-2x^2)+150x+C

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integral of 0.024(0.06x^2-4x)+150

\int 0.024 (0.06 x^{2} - 4 x) + 150 d x

0.024 (0.02 x^{3} - 2 x^{2}) + 150 x + C

Solution steps

\int 0.024 (0.06 x^{2} - 4 x) + 150 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 0.024 (0.06 x^{2} - 4 x) d x + \int 150 d x

\int 0.024 (0.06 x^{2} - 4 x) d x = 0.024 (0.02 x^{3} - 2 x^{2})

\int 150 d x = 150 x

= 0.024 (0.02 x^{3} - 2 x^{2}) + 150 x

Add a constant to the solution

= 0.024 (0.02 x^{3} - 2 x^{2}) + 150 x + C

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What is the integral of 0.024(0.06x^2-4x)+150 ?
The integral of 0.024(0.06x^2-4x)+150 is 0.024(0.02x^3-2x^2)+150x+C