What is the integral from 0 to 6 of (x-4)^{-2/3} ?

The integral from 0 to 6 of (x-4)^{-2/3} is 3*4^{1/3}+3*2^{1/3}

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integral from 0 to 6 of (x-4)^{-2/3}

\int_{0}^{6} (x - 4)^{- \frac{2}{3}} d x

3 \cdot 4^{\frac{1}{3}} + 3 \cdot 2^{\frac{1}{3}}

Decimal

8.54196 \dots

Solution steps

\int_{0}^{6} (x - 4)^{- \frac{2}{3}} d x

Find undefined (singularity) points:

x = 4

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{0}^{4} (x - 4)^{- \frac{2}{3}} d x + \int_{4}^{6} (x - 4)^{- \frac{2}{3}} d x

\int_{0}^{4} (x - 4)^{- \frac{2}{3}} d x = - 3 (- 1)^{\frac{1}{3}} \cdot 4^{\frac{1}{3}}

\int_{4}^{6} (x - 4)^{- \frac{2}{3}} d x = 3 \cdot 2^{\frac{1}{3}}

= - 3 (- 1)^{\frac{1}{3}} \cdot 4^{\frac{1}{3}} + 3 \cdot 2^{\frac{1}{3}}

Simplify

- 3 (- 1)^{\frac{1}{3}} \cdot 4^{\frac{1}{3}} + 3 \cdot 2^{\frac{1}{3}} : 3 \cdot 4^{\frac{1}{3}} + 3 \cdot 2^{\frac{1}{3}}

= 3 \cdot 4^{\frac{1}{3}} + 3 \cdot 2^{\frac{1}{3}}

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What is the integral from 0 to 6 of (x-4)^{-2/3} ?
The integral from 0 to 6 of (x-4)^{-2/3} is 3*4^{1/3}+3*2^{1/3}