What is the integral of (3x^2)/2-x-16.5 ?

The integral of (3x^2)/2-x-16.5 is 0.5x^3-0.5x^2-16.5x+C

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integral of (3x^2)/2-x-16.5

\int \frac{3 x ^{2}}{2} - x - 16.5 d x

0.5 x^{3} - 0.5 x^{2} - 16.5 x + C

Solution steps

\int \frac{3 x ^{2}}{2} - x - 16.5 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 \frac{3 x ^{2}}{2} d x - \int x d x - \int 16.5 d x

\int \frac{3 x ^{2}}{2} d x = \frac{x ^{3}}{2}

\int x d x = \frac{x ^{2}}{2}

\int 16.5 d x = 16.5 x

= \frac{x ^{3}}{2} - \frac{x ^{2}}{2} - 16.5 x

Simplify

= 0.5 x^{3} - 0.5 x^{2} - 16.5 x

Add a constant to the solution

= 0.5 x^{3} - 0.5 x^{2} - 16.5 x + C

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What is the integral of (3x^2)/2-x-16.5 ?
The integral of (3x^2)/2-x-16.5 is 0.5x^3-0.5x^2-16.5x+C