What is the integral from 4y^2+4z^2 to 4 of x ?

The integral from 4y^2+4z^2 to 4 of x is 8-((4z^2+4y^2)^2)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 4y^2+4z^2 to 4 of x

\int_{4 y^{2} + 4 z^{2}}^{4} x d x

8 - \frac{( 4 z ^{2} + 4 y ^{2} ) ^{2}}{2}

Solution steps

\int_{4 y^{2} + 4 z^{2}}^{4} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{4 z^{2} + 4 y^{2}}^{4}

= [\frac{x ^{2}}{2}]_{4 z^{2} + 4 y^{2}}^{4}

Compute the boundaries:

8 - \frac{( 4 z ^{2} + 4 y ^{2} ) ^{2}}{2}

= 8 - \frac{( 4 z ^{2} + 4 y ^{2} ) ^{2}}{2}

\frac{d}{d x} ((x^{2} - 1) (x^{3} + 3 x))

(y^{2})^{^{'}}

f (x) = x^{2} - x + 2

\frac{d y}{d x} y = x^{- 12}

\frac{\partial}{\partial y} (x^{2} y + e^{x y})

What is the integral from 4y^2+4z^2 to 4 of x ?
The integral from 4y^2+4z^2 to 4 of x is 8-((4z^2+4y^2)^2)/2