What is the integral of (1-y)^3 ?

The integral of (1-y)^3 is y-(3y^2)/2+y^3-(y^4)/4+C

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integral of (1-y)^3

\int (1 - y)^{3} d y

y - \frac{3 y ^{2}}{2} + y^{3} - \frac{y ^{4}}{4} + C

Solution steps

\int (1 - y)^{3} d y

Expand

(1 - y)^{3} : 1 - 3 y + 3 y^{2} - y^{3}

= \int 1 - 3 y + 3 y^{2} - 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 1 d y - \int 3 y d y + \int 3 y^{2} d y - \int y^{3} d y

\int 1 d y = y

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

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

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

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

Add a constant to the solution

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

\int y cos (4 y) d y

y = \frac{( 2 x ^{2} - 1 ) ^{3}}{( 5 x - 8 ) ^{2}}

\frac{\partial}{\partial x} (x x + 1)

\frac{d y}{d x} = 2 y

\int_{0}^{6} 3 x - x^{2} d x

What is the integral of (1-y)^3 ?
The integral of (1-y)^3 is y-(3y^2)/2+y^3-(y^4)/4+C