What is the integral of 5-0.03x+0.0004x^2 ?

The integral of 5-0.03x+0.0004x^2 is 5x-0.015x^2+0.00013…x^3+C

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integral of 5-0.03x+0.0004x^2

\int 5 - 0.03 x + 0.0004 x^{2} d x

5 x - 0.015 x^{2} + 0.00013 \dots x^{3} + C

Solution steps

\int 5 - 0.03 x + 0.0004 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 5 d x - \int 0.03 x d x + \int 0.0004 x^{2} d x

\int 5 d x = 5 x

\int 0.03 x d x = 0.015 x^{2}

\int 0.0004 x^{2} d x = 0.00013 \dots x^{3}

= 5 x - 0.015 x^{2} + 0.00013 \dots x^{3}

Add a constant to the solution

= 5 x - 0.015 x^{2} + 0.00013 \dots x^{3} + C

What is the integral of 5-0.03x+0.0004x^2 ?
The integral of 5-0.03x+0.0004x^2 is 5x-0.015x^2+0.00013…x^3+C