What is the integral from 0 to 1-x of 24y(1-x-y) ?

The integral from 0 to 1-x of 24y(1-x-y) is 4(-x^3+3x^2-3x+1)

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integral from 0 to 1-x of 24y(1-x-y)

\int_{0}^{1 - x} 24 y (1 - x - y) d y

4 (- x^{3} + 3 x^{2} - 3 x + 1)

Solution steps

\int_{0}^{1 - x} 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_{0}^{1 - x} (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]_{0}^{1 - x}

= 24 [\frac{1}{2} y^{2} (1 - x - y) - \int - \frac{y ^{2}}{2} d y]_{0}^{1 - x}

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

= 24 [\frac{1}{2} y^{2} (1 - x - y) - (- \frac{y ^{3}}{6})]_{0}^{1 - x}

Simplify

24 [\frac{1}{2} y^{2} (1 - x - y) - (- \frac{y ^{3}}{6})]_{0}^{1 - x} : 24 [\frac{1}{6} (3 y^{2} - 3 x y^{2} - 2 y^{3})]_{0}^{1 - x}

= 24 [\frac{1}{6} (3 y^{2} - 3 x y^{2} - 2 y^{3})]_{0}^{1 - x}

Compute the boundaries:

\frac{- 3 x ^{3} + 9 x ^{2} - 9 x - 2 ( - x + 1 ) ^{3} + 3}{6}

= 24 \cdot \frac{- 3 x ^{3} + 9 x ^{2} - 9 x - 2 ( - x + 1 ) ^{3} + 3}{6}

Simplify

= 4 (- x^{3} + 3 x^{2} - 3 x + 1)

4 y^{2}

\int x \cdot sin^{3} (x) d x

y = \frac{6 x}{x + 3}, (- 1, - 3)

f (x) = \frac{2 x}{x ^{2} - 8}

\int \frac{1}{t ^{2} - 2 t + 5} d t

What is the integral from 0 to 1-x of 24y(1-x-y) ?
The integral from 0 to 1-x of 24y(1-x-y) is 4(-x^3+3x^2-3x+1)