What is the integral from 1/2 to 2 of \sqrt[3]{x} ?

The integral from 1/2 to 2 of \sqrt[3]{x} is 3/(2^{2/3)}-3/(8\sqrt[3]{2)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1/2 to 2 of \sqrt[3]{x}

\int_{\frac{1}{2}}^{2} 3 x d x

\frac{3}{2 ^{\frac{2}{3}}} - \frac{3}{8 3 2}

Decimal

1.59224 \dots

Solution steps

\int_{\frac{1}{2}}^{2} 3 x d x

Apply the Power Rule

= [\frac{3}{4} x^{\frac{4}{3}}]_{\frac{1}{2}}^{2}

Compute the boundaries:

\frac{3}{2 ^{\frac{2}{3}}} - \frac{3}{8 3 2}

= \frac{3}{2 ^{\frac{2}{3}}} - \frac{3}{8 3 2}

\int_{0}^{4} π (4 x)^{2} d x

\int_{0}^{1} (3 x^{3} - x^{2} + 2 x - 4) d x

\int_{0}^{1} \frac{32}{9} (x y)^{\frac{1}{3}} d y

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

\int_{0}^{5} (x + 6 - x^{2}) d x

What is the integral from 1/2 to 2 of \sqrt[3]{x} ?
The integral from 1/2 to 2 of \sqrt[3]{x} is 3/(2^{2/3)}-3/(8\sqrt[3]{2)}