What is the integral from 0 to 1/2-x of 384xy ?

The integral from 0 to 1/2-x of 384xy is 48x(-2x+1)^2

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integral from 0 to 1/2-x of 384xy

\int_{0}^{\frac{1}{2} - x} 384 x y d y

48 x (- 2 x + 1)^{2}

Solution steps

\int_{0}^{\frac{1}{2} - x} 384 x y d y

Take the constant out:

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

= 384 x \cdot \int_{0}^{\frac{1}{2} - x} y d y

Apply the Power Rule

: [\frac{y ^{2}}{2}]_{0}^{\frac{1}{2} - x}

= 384 x [\frac{y ^{2}}{2}]_{0}^{\frac{1}{2} - x}

Compute the boundaries:

\frac{( - 2 x + 1 ) ^{2}}{8}

= 384 x \frac{( - 2 x + 1 ) ^{2}}{8}

Simplify

= 48 x (- 2 x + 1)^{2}

\int x (4 x^{2} + 7)^{9} d x

\frac{d y}{d x} = \frac{y}{1 + x ^{2}}

f^{^{''}} (x) = \frac{3}{4} x^{4} + 2 x^{3} - 12 x^{2} + 29

\frac{\partial}{\partial y} (x + 4 y)

What is the integral from 0 to 1/2-x of 384xy ?
The integral from 0 to 1/2-x of 384xy is 48x(-2x+1)^2