What is the integral from 0 to 1-y of 6x ?

The integral from 0 to 1-y of 6x is 3(-y+1)^2

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integral from 0 to 1-y of 6x

\int_{0}^{1 - y} 6 x d x

3 (- y + 1)^{2}

Solution steps

\int_{0}^{1 - y} 6 x d x

Take the constant out:

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

= 6 \cdot \int_{0}^{1 - y} x d x

Apply the Power Rule

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

Compute the boundaries:

\frac{( - y + 1 ) ^{2}}{2}

= 6 \cdot \frac{( - y + 1 ) ^{2}}{2}

Simplify

= 3 (- y + 1)^{2}

\frac{\partial}{\partial x} (3 (\frac{( 4 + x )}{5 + z})^{2})

\int (sin (x) + sinh (x)) d x

in v erse l a pl a ce \frac{3 s + 12}{s ( s ^{2} + 4 )}

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

\int \frac{3}{x ^{2} + 4} d x

What is the integral from 0 to 1-y of 6x ?
The integral from 0 to 1-y of 6x is 3(-y+1)^2