What is the integral of 5-0.04x+3x^2 ?

The integral of 5-0.04x+3x^2 is 5x-0.02x^2+x^3+C

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integral of 5-0.04x+3x^2

\int 5 - 0.04 x + 3 x^{2} d x

5 x - 0.02 x^{2} + x^{3} + C

Solution steps

\int 5 - 0.04 x + 3 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 0.04 x d x + \int 3 x^{2} d x

\int 5 d x = 5 x

\int 0.04 x d x = 0.02 x^{2}

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

= 5 x - 0.02 x^{2} + x^{3}

Add a constant to the solution

= 5 x - 0.02 x^{2} + x^{3} + C

\frac{d}{d x} (\frac{2 x}{x ^{2} - 36})

4 y^{^{''}} - 4 y^{^{'}} - 3 y = 0, y (0) = 1, y^{^{'}} (0) = 9

x \to - π lim (csc (x))

\frac{\partial}{\partial x} ((e^{x} + e^{y})^{6})

\int \frac{1 - x ^{2}}{( x ^{2} - 1 ) ^{2}} d x

What is the integral of 5-0.04x+3x^2 ?
The integral of 5-0.04x+3x^2 is 5x-0.02x^2+x^3+C