What is the integral from-1 to 1 of x^2y^2z^2 ?

The integral from-1 to 1 of x^2y^2z^2 is (2y^2z^2)/3

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integral from-1 to 1 of x^2y^2z^2

\int_{- 1}^{1} x^{2} y^{2} z^{2} d x

\frac{2 y ^{2} z ^{2}}{3}

Solution steps

\int_{- 1}^{1} x^{2} y^{2} z^{2} d x

Take the constant out:

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

= y^{2} z^{2} \cdot \int_{- 1}^{1} x^{2} d x

Apply the Power Rule

: [\frac{x ^{3}}{3}]_{- 1}^{1}

= y^{2} z^{2} [\frac{x ^{3}}{3}]_{- 1}^{1}

Compute the boundaries:

\frac{2}{3}

= y^{2} z^{2} \frac{2}{3}

Simplify

= \frac{2 y ^{2} z ^{2}}{3}

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What is the integral from-1 to 1 of x^2y^2z^2 ?
The integral from-1 to 1 of x^2y^2z^2 is (2y^2z^2)/3