What is the integral of (y^3+1.8y^2-2.4y) ?

The integral of (y^3+1.8y^2-2.4y) is 0.25y^4+0.6y^3-1.2y^2+C

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integral of (y^3+1.8y^2-2.4y)

\int (y^{3} + 1.8 y^{2} - 2.4 y) d y

0.25 y^{4} + 0.6 y^{3} - 1.2 y^{2} + C

Solution steps

\int y^{3} + 1.8 y^{2} - 2.4 y d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int y^{3} d y + \int 1.8 y^{2} d y - \int 2.4 y d y

\int y^{3} d y = \frac{y ^{4}}{4}

\int 1.8 y^{2} d y = 0.6 y^{3}

\int 2.4 y d y = 1.2 y^{2}

= \frac{y ^{4}}{4} + 0.6 y^{3} - 1.2 y^{2}

Simplify

= 0.25 y^{4} + 0.6 y^{3} - 1.2 y^{2}

Add a constant to the solution

= 0.25 y^{4} + 0.6 y^{3} - 1.2 y^{2} + C

x \to - 2 lim (\frac{3 x - 1}{x})

\frac{\partial}{\partial x} (\frac{4 x}{x ^{2} + 2})

(0, 4), (1, 1)

e^{- x}, 4, 0.25

\frac{\partial}{\partial x} (x z + x yz)

What is the integral of (y^3+1.8y^2-2.4y) ?
The integral of (y^3+1.8y^2-2.4y) is 0.25y^4+0.6y^3-1.2y^2+C