What is the integral from 0 to 7 of 1/((6-2x)^{1/3)} ?

The integral from 0 to 7 of 1/((6-2x)^{1/3)} is (3*3^{2/3})/(2*2^{1/3)}-3

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integral from 0 to 7 of 1/((6-2x)^{1/3)}

\int_{0}^{7} \frac{1}{( 6 - 2 x ) ^{\frac{1}{3}}} d x

\frac{3 \cdot 3 ^{\frac{2}{3}}}{2 \cdot 2 ^{\frac{1}{3}}} - 3

Decimal

- 0.52355 \dots

Solution steps

\int_{0}^{7} \frac{1}{( 6 - 2 x ) ^{\frac{1}{3}}} d x

Find undefined (singularity) points:

x = 3

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}^{3} \frac{1}{( 6 - 2 x ) ^{\frac{1}{3}}} d x + \int_{3}^{7} \frac{1}{( 6 - 2 x ) ^{\frac{1}{3}}} d x

\int_{0}^{3} \frac{1}{( 6 - 2 x ) ^{\frac{1}{3}}} d x = \frac{3 \cdot 3 ^{\frac{2}{3}}}{2 \cdot 2 ^{\frac{1}{3}}}

\int_{3}^{7} \frac{1}{( 6 - 2 x ) ^{\frac{1}{3}}} d x = - 3 (- 1)^{\frac{2}{3}}

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

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

= \frac{3 \cdot 3 ^{\frac{2}{3}}}{2 \cdot 2 ^{\frac{1}{3}}} - 3

What is the integral from 0 to 7 of 1/((6-2x)^{1/3)} ?
The integral from 0 to 7 of 1/((6-2x)^{1/3)} is (3*3^{2/3})/(2*2^{1/3)}-3