What is the integral from 0 to 8 of (\sqrt[3]{x})^2 ?

The integral from 0 to 8 of (\sqrt[3]{x})^2 is 96/5

Popular >

integral from 0 to 8 of (\sqrt[3]{x})^2

\int_{0}^{8} (3 x)^{2} d x

\frac{96}{5}

Decimal

19.2

Solution steps

\int_{0}^{8} (3 x)^{2} d x

Apply Integration By Parts

: [x (3 x)^{2} - \int \frac{2}{3} x^{\frac{2}{3}} d x]_{0}^{8}

= [x (3 x)^{2} - \int \frac{2}{3} x^{\frac{2}{3}} d x]_{0}^{8}

\int \frac{2}{3} x^{\frac{2}{3}} d x = \frac{2}{5} x^{\frac{5}{3}}

= [x (3 x)^{2} - \frac{2}{5} x^{\frac{5}{3}}]_{0}^{8}

Simplify

[x (3 x)^{2} - \frac{2}{5} x^{\frac{5}{3}}]_{0}^{8} : [\frac{3}{5} x^{\frac{5}{3}}]_{0}^{8}

= [\frac{3}{5} x^{\frac{5}{3}}]_{0}^{8}

Compute the boundaries:

\frac{96}{5}

= \frac{96}{5}

\int_{0}^{π} sin^{3} (φ) d φ

\int_{0}^{\frac{2}{3}} 1 + 36 x d x

\int_{- 3}^{3} x^{3} + \frac{1}{x} d x

\int_{0}^{\frac{1}{4}} y arctan (4 y) d y

\int_{0}^{\frac{π}{2}} sin^{2} (2 n x) d x

What is the integral from 0 to 8 of (\sqrt[3]{x})^2 ?
The integral from 0 to 8 of (\sqrt[3]{x})^2 is 96/5