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

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

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integral of (1-x-y)^3

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

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

Solution steps

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

Expand

(1 - x - y)^{3} : - y^{3} - 3 x y^{2} + 3 y^{2} - 3 x^{2} y + 6 x y - 3 y - x^{3} + 3 x^{2} - 3 x + 1

= \int - y^{3} - 3 x y^{2} + 3 y^{2} - 3 x^{2} y + 6 x y - 3 y - x^{3} + 3 x^{2} - 3 x + 1 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 3 x y^{2} d y + \int 3 y^{2} d y - \int 3 x^{2} y d y + \int 6 x y d y - \int 3 y d y - \int x^{3} d y + \int 3 x^{2} d y - \int 3 x d y + \int 1 d y

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

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

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

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

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

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

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

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

\int 3 x d y = 3 x y

\int 1 d y = y

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

Add a constant to the solution

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

(3, - 5), (5, - 5)

\int x^{2} ln^{2} (x) d x

y^{^{'}} + 3 x^{2} y = x e^{- x^{3}}

y^{^{'}} = \frac{3 sec ( y )}{( x + 5 ) ^{2}}

f (x) = 2^{30}

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