What is the integral from 0 to 1 of \sqrt[3]{t} ?

The integral from 0 to 1 of \sqrt[3]{t} is 3/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of \sqrt[3]{t}

\frac{3}{4}

Solution steps

Apply the Power Rule

: [\frac{3}{4} t^{\frac{4}{3}}]_{0}^{1}

= [\frac{3}{4} t^{\frac{4}{3}}]_{0}^{1}

Compute the boundaries:

\frac{3}{4}

= \frac{3}{4}

\int_{0}^{3} (5 x - 4) d x

\int_{0}^{1} 3 t + 1 d t

\int_{1}^{7} 2 π y (7 - y) d y

\int_{0}^{\frac{π}{2}} e^{- 6 x} cos (x) d x

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

What is the integral from 0 to 1 of \sqrt[3]{t} ?
The integral from 0 to 1 of \sqrt[3]{t} is 3/4