What is the integral of-2x+18x^2-4x^3+16 ?

The integral of-2x+18x^2-4x^3+16 is -x^2+6x^3-x^4+16x+C

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integral of-2x+18x^2-4x^3+16

\int - 2 x + 18 x^{2} - 4 x^{3} + 16 d x

- x^{2} + 6 x^{3} - x^{4} + 16 x + C

Solution steps

\int - 2 x + 18 x^{2} - 4 x^{3} + 16 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 2 x d x + \int 18 x^{2} d x - \int 4 x^{3} d x + \int 16 d x

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

\int 18 x^{2} d x = 6 x^{3}

\int 4 x^{3} d x = x^{4}

\int 16 d x = 16 x

= - x^{2} + 6 x^{3} - x^{4} + 16 x

Add a constant to the solution

= - x^{2} + 6 x^{3} - x^{4} + 16 x + C

x \to 3 lim (4 x^{2} - 2 x - 6)

y = \frac{x}{x + 9}, (1, 0.1)

x \to \infty lim (- 5 x^{2})

f (x) = \frac{5}{x ^{2}} + sin (x)

x^{^{'}} = 0.1078 \cdot (1 - \frac{x}{105900}) x - x

What is the integral of-2x+18x^2-4x^3+16 ?
The integral of-2x+18x^2-4x^3+16 is -x^2+6x^3-x^4+16x+C