What is the integral from 8y^2+8z^2 to 8 of x ?

The integral from 8y^2+8z^2 to 8 of x is 32-((8z^2+8y^2)^2)/2

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integral from 8y^2+8z^2 to 8 of x

\int_{8 y^{2} + 8 z^{2}}^{8} x d x

32 - \frac{( 8 z ^{2} + 8 y ^{2} ) ^{2}}{2}

Solution steps

\int_{8 y^{2} + 8 z^{2}}^{8} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{8 z^{2} + 8 y^{2}}^{8}

= [\frac{x ^{2}}{2}]_{8 z^{2} + 8 y^{2}}^{8}

Compute the boundaries:

32 - \frac{( 8 z ^{2} + 8 y ^{2} ) ^{2}}{2}

= 32 - \frac{( 8 z ^{2} + 8 y ^{2} ) ^{2}}{2}

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What is the integral from 8y^2+8z^2 to 8 of x ?
The integral from 8y^2+8z^2 to 8 of x is 32-((8z^2+8y^2)^2)/2