What is the integral from 0 to 8 of 2bx^{-1/3} ?

The integral from 0 to 8 of 2bx^{-1/3} is 12b

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integral from 0 to 8 of 2bx^{-1/3}

\int_{0}^{8} 2 b x^{- \frac{1}{3}} d x

12 b

Solution steps

\int_{0}^{8} 2 b x^{- \frac{1}{3}} d x

Take the constant out:

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

= 2 b \cdot \int_{0}^{8} x^{- \frac{1}{3}} d x

Apply the Power Rule

: [\frac{3}{2} x^{\frac{2}{3}}]_{0}^{8}

= 2 b [\frac{3}{2} x^{\frac{2}{3}}]_{0}^{8}

Compute the boundaries:

6

= 2 b \cdot 6

Simplify

= 12 b

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What is the integral from 0 to 8 of 2bx^{-1/3} ?
The integral from 0 to 8 of 2bx^{-1/3} is 12b