What is the integral from z to sqrt(4-z^2 of)z ?

The integral from z to sqrt(4-z^2 of)z is 2-z^2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from z to sqrt(4-z^2 of)z

\int_{z}^{4 - z^{2}} z d z

2 - z^{2}

Solution steps

\int_{z}^{4 - z^{2}} z d z

Apply the Power Rule

: [\frac{z ^{2}}{2}]_{z}^{4 - z^{2}}

= [\frac{z ^{2}}{2}]_{z}^{4 - z^{2}}

Compute the boundaries:

\frac{1}{2} (4 - z^{2}) - \frac{z ^{2}}{2}

= \frac{1}{2} (4 - z^{2}) - \frac{z ^{2}}{2}

Simplify

= 2 - z^{2}

\int_{c o s (x)}^{s i n (x)} ln (1 + 2 v) d v

\int_{0}^{1} (x^{18} + 1 8^{x}) d x

\int_{0}^{\infty} 2 e^{- x - y} d y

\int_{0}^{1} ln (x^{2}) d x

\int_{0}^{3} \frac{x}{x ^{2} + 7} d x

What is the integral from z to sqrt(4-z^2 of)z ?
The integral from z to sqrt(4-z^2 of)z is 2-z^2