What is the integral of 24y(1-x-y) ?

The integral of 24y(1-x-y) is 24(1/2 y^2(1-x-y)+(y^3)/6)+C

Popular >

integral of 24y(1-x-y)

\int 24 y (1 - x - y) d y

24 (\frac{1}{2} y^{2} (1 - x - y) + \frac{y ^{3}}{6}) + C

Solution steps

\int 24 y (1 - x - y) d y

Take the constant out:

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

= 24 \cdot \int (1 - x - y) y d y

Apply Integration By Parts

: \frac{1}{2} y^{2} (1 - x - y) - \int - \frac{y ^{2}}{2} d y

= 24 (\frac{1}{2} y^{2} (1 - x - y) - \int - \frac{y ^{2}}{2} d y)

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

= 24 (\frac{1}{2} y^{2} (1 - x - y) - (- \frac{y ^{3}}{6}))

Simplify

= 24 (\frac{1}{2} y^{2} (1 - x - y) + \frac{y ^{3}}{6})

Add a constant to the solution

= 24 (\frac{1}{2} y^{2} (1 - x - y) + \frac{y ^{3}}{6}) + C

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

\frac{d}{d x} (\frac{( 3 x + 1 )}{x ^{\frac{2}{3}}})

e^{- 2 s} (\frac{1}{s + 1})

\int t e^{- 2 t} d t

\frac{\partial}{\partial x} (50 x - 0.5 x^{2} + 75 y - y^{2})

What is the integral of 24y(1-x-y) ?
The integral of 24y(1-x-y) is 24(1/2 y^2(1-x-y)+(y^3)/6)+C