What is the integral from 0 to t^8 of sqrt(u) ?

The integral from 0 to t^8 of sqrt(u) is (2t^{12})/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to t^8 of sqrt(u)

\int_{0}^{t^{8}} u d u

\frac{2 t ^{12}}{3}

Solution steps

\int_{0}^{t^{8}} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{t^{8}}

= [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{t^{8}}

Compute the boundaries:

\frac{2}{3} t^{12}

= \frac{2}{3} t^{12}

Simplify

= \frac{2 t ^{12}}{3}

\int_{0}^{8.9} 960 t e^{- s t} d t

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

\int_{2.6}^{3} \frac{2 x}{9} d x

\int_{- \frac{π}{2}}^{\frac{π}{2}} x cos (x^{2}) d x

\int_{1}^{7} \frac{x}{7} - \frac{7}{x} d x

What is the integral from 0 to t^8 of sqrt(u) ?
The integral from 0 to t^8 of sqrt(u) is (2t^{12})/3