What is the integral from x to 1 of 6xyz ?

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

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integral from x to 1 of 6xyz

\int_{x}^{1} 6 x yz d z

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

Solution steps

\int_{x}^{1} 6 x yz d z

Take the constant out:

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

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

Apply the Power Rule

: [\frac{z ^{2}}{2}]_{x}^{1}

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

Compute the boundaries:

\frac{1}{2} - \frac{x ^{2}}{2}

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

Simplify

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

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