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

The integral from 2 to 4 of x^{1/3}(1-2x) is 3*4^{1/3}+(27-192*2^{1/3})/(7*2^{2/3)}

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integral from 2 to 4 of x^{1/3}(1-2x)

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

3 \cdot 4^{\frac{1}{3}} + \frac{27 - 192 \cdot 2 ^{\frac{1}{3}}}{7 \cdot 2 ^{\frac{2}{3}}}

Decimal Notation

- 14.57802 \dots

Solution steps

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

Expand

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

= \int_{2}^{4} x^{\frac{1}{3}} - 2 x^{\frac{4}{3}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{2}^{4} x^{\frac{1}{3}} d x - \int_{2}^{4} 2 x^{\frac{4}{3}} d x

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

\int_{2}^{4} 2 x^{\frac{4}{3}} d x = \frac{2 ( 48 \cdot 4 ^{\frac{1}{3}} - 12 \cdot 2 ^{\frac{1}{3}} )}{7}

= 3 \cdot 4^{\frac{1}{3}} - \frac{3}{2 ^{\frac{2}{3}}} - \frac{2 ( 48 \cdot 4 ^{\frac{1}{3}} - 12 \cdot 2 ^{\frac{1}{3}} )}{7}

Simplify

3 \cdot 4^{\frac{1}{3}} - \frac{3}{2 ^{\frac{2}{3}}} - \frac{2 ( 48 \cdot 4 ^{\frac{1}{3}} - 12 \cdot 2 ^{\frac{1}{3}} )}{7} : 3 \cdot 4^{\frac{1}{3}} + \frac{27 - 192 \cdot 2 ^{\frac{1}{3}}}{7 \cdot 2 ^{\frac{2}{3}}}

= 3 \cdot 4^{\frac{1}{3}} + \frac{27 - 192 \cdot 2 ^{\frac{1}{3}}}{7 \cdot 2 ^{\frac{2}{3}}}

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What is the integral from 2 to 4 of x^{1/3}(1-2x) ?
The integral from 2 to 4 of x^{1/3}(1-2x) is 3*4^{1/3}+(27-192*2^{1/3})/(7*2^{2/3)}