What is the integral of 6y(x-y) ?

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

integral of 6y(x-y)

\int 6 y (x - y) d y

6 (\frac{x y ^{2}}{2} - \frac{y ^{3}}{3}) + C

Solution steps

\int 6 y (x - y) d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int (x - y) y d y

Expand

(x - y) y : x y - y^{2}

= 6 \cdot \int x y - y^{2} 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

= 6 (\int x y d y - \int y^{2} d y)

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

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

= 6 (\frac{x y ^{2}}{2} - \frac{y ^{3}}{3})

Add a constant to the solution

= 6 (\frac{x y ^{2}}{2} - \frac{y ^{3}}{3}) + C

x \to 0 lim (\frac{2 x + 4 - 2}{x + 1 - 1})

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

\frac{d}{d x} (e^{- x} - y)

t an g e n t f (x) = 4 x^{2} - 5 x, (- 1, 9)

n = 1 \sum \infty \frac{3 ^{n}}{n ^{3}}

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