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

The integral of y^3(2y^2-3) is (y^6)/3-(3y^4)/4+C

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integral of y^3(2y^2-3)

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

\frac{y ^{6}}{3} - \frac{3 y ^{4}}{4} + C

Solution steps

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

Expand

y^{3} (2 y^{2} - 3) : 2 y^{5} - 3 y^{3}

= \int 2 y^{5} - 3 y^{3} 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 2 y^{5} d y - \int 3 y^{3} d y

\int 2 y^{5} d y = \frac{y ^{6}}{3}

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

= \frac{y ^{6}}{3} - \frac{3 y ^{4}}{4}

Add a constant to the solution

= \frac{y ^{6}}{3} - \frac{3 y ^{4}}{4} + C

\frac{\partial}{\partial x} (2 y + y e^{x})

\frac{d x}{d t} = - 3 x

h \to 0 lim (\frac{e ^{2 h} - 1}{h})

\frac{d}{d x} (x \cdot cos (x))

f (x) = 2 x 5 - 2 x, at x = - 1

What is the integral of y^3(2y^2-3) ?
The integral of y^3(2y^2-3) is (y^6)/3-(3y^4)/4+C