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

The integral from-6 to-1 of 2x^{1/3} is 2(-(3(-1)^{1/3})/4+(9(-1)^{1/3}*3^{1/3})/(2^{2/3)})

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integral from-6 to-1 of 2x^{1/3}

\int_{- 6}^{- 1} 2 x^{\frac{1}{3}} d x

2 (- \frac{3 ( - 1 ) ^{\frac{1}{3}}}{4} + \frac{9 ( - 1 ) ^{\frac{1}{3}} \cdot 3 ^{\frac{1}{3}}}{2 ^{\frac{2}{3}}})

Solution steps

\int_{- 6}^{- 1} 2 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 \cdot \int_{- 6}^{- 1} x^{\frac{1}{3}} d x

Apply the Power Rule

= 2 [\frac{3}{4} x^{\frac{4}{3}}]_{- 6}^{- 1}

Compute the boundaries:

- (- 1)^{\frac{1}{3}} \frac{3}{4} + \frac{9 ( - 1 ) ^{\frac{1}{3}} \cdot 3 ^{\frac{1}{3}}}{2 ^{\frac{2}{3}}}

= 2 (- (- 1)^{\frac{1}{3}} \frac{3}{4} + \frac{9 ( - 1 ) ^{\frac{1}{3}} \cdot 3 ^{\frac{1}{3}}}{2 ^{\frac{2}{3}}})

Simplify

= 2 (- \frac{3 ( - 1 ) ^{\frac{1}{3}}}{4} + \frac{9 ( - 1 ) ^{\frac{1}{3}} \cdot 3 ^{\frac{1}{3}}}{2 ^{\frac{2}{3}}})

What is the integral from-6 to-1 of 2x^{1/3} ?
The integral from-6 to-1 of 2x^{1/3} is 2(-(3(-1)^{1/3})/4+(9(-1)^{1/3}*3^{1/3})/(2^{2/3)})