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

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

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

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

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

Solution steps

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

Apply the Power Rule

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

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

Compute the boundaries:

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

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

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

\int x (2 + x x)^{5} d x

\frac{d}{d x} (- e^{- x} x^{2} + e^{- x} x)

e^{- x}, 5

\frac{d y}{d x} = x y

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