What is the integral from 0 to 10 of 9y^{1/3} ?

The integral from 0 to 10 of 9y^{1/3} is (135*5^{1/3})/(2^{2/3)}

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integral from 0 to 10 of 9y^{1/3}

\int_{0}^{10} 9 y^{\frac{1}{3}} d y

\frac{135 \cdot 5 ^{\frac{1}{3}}}{2 ^{\frac{2}{3}}}

Decimal

145.42434 \dots

Solution steps

\int_{0}^{10} 9 y^{\frac{1}{3}} d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int_{0}^{10} y^{\frac{1}{3}} d y

Apply the Power Rule

= 9 [\frac{3}{4} y^{\frac{4}{3}}]_{0}^{10}

Compute the boundaries:

\frac{15 \cdot 5 ^{\frac{1}{3}}}{2 ^{\frac{2}{3}}}

= 9 \cdot \frac{15 \cdot 5 ^{\frac{1}{3}}}{2 ^{\frac{2}{3}}}

Simplify

= \frac{135 \cdot 5 ^{\frac{1}{3}}}{2 ^{\frac{2}{3}}}

What is the integral from 0 to 10 of 9y^{1/3} ?
The integral from 0 to 10 of 9y^{1/3} is (135*5^{1/3})/(2^{2/3)}