What is the integral from 0 to 1 of (\sqrt[3]{y})^2 ?

The integral from 0 to 1 of (\sqrt[3]{y})^2 is 3/5

Popular >

integral from 0 to 1 of (\sqrt[3]{y})^2

\int_{0}^{1} (3 y)^{2} d y

\frac{3}{5}

Decimal

0.6

Solution steps

\int_{0}^{1} (3 y)^{2} d y

Apply Integration By Parts

: [y (3 y)^{2} - \int \frac{2}{3} y^{\frac{2}{3}} d y]_{0}^{1}

= [y (3 y)^{2} - \int \frac{2}{3} y^{\frac{2}{3}} d y]_{0}^{1}

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

= [y (3 y)^{2} - \frac{2}{5} y^{\frac{5}{3}}]_{0}^{1}

Simplify

[y (3 y)^{2} - \frac{2}{5} y^{\frac{5}{3}}]_{0}^{1} : [\frac{3}{5} y^{\frac{5}{3}}]_{0}^{1}

= [\frac{3}{5} y^{\frac{5}{3}}]_{0}^{1}

Compute the boundaries:

\frac{3}{5}

= \frac{3}{5}

\int_{- 1}^{3} 6 x y - 5 e^{y} d x

\int_{0}^{4} \frac{x}{4 + x ^{2}} d x

\int_{- 2}^{0} x^{3} + 3 x^{2} + 2 x d x

\int_{1}^{\infty} e^{- x + 1} d x

\int_{0}^{4} \frac{x ^{2}}{x ^{3} + 8} d x

What is the integral from 0 to 1 of (\sqrt[3]{y})^2 ?
The integral from 0 to 1 of (\sqrt[3]{y})^2 is 3/5