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

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

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

\int_{0}^{1} 3 y d y

\frac{3}{4}

Decimal

0.75

Solution steps

\int_{0}^{1} 3 y d y

Apply the Power Rule

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

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

Compute the boundaries:

\frac{3}{4}

= \frac{3}{4}

\int_{0}^{3} 4 x - x^{2} - x d x

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

\int_{0}^{\infty} e^{- a x} x d x

\int_{0}^{5} (2 x + 3) d x

\int_{0}^{1} 16 t^{3} d t

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