What is the integral from-3 to-2 of 5(2x+4)^{1/3} ?

The integral from-3 to-2 of 5(2x+4)^{1/3} is (15(-1)^{1/3})/(2*2^{2/3)}

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integral from-3 to-2 of 5(2x+4)^{1/3}

\int_{- 3}^{- 2} 5 (2 x + 4)^{\frac{1}{3}} d x

\frac{15 ( - 1 ) ^{\frac{1}{3}}}{2 \cdot 2 ^{\frac{2}{3}}}

Solution steps

\int_{- 3}^{- 2} 5 (2 x + 4)^{\frac{1}{3}} d x

Take the constant out:

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

= 5 \cdot \int_{- 3}^{- 2} (2 x + 4)^{\frac{1}{3}} d x

Apply u-substitution

: \int_{- 2}^{0} \frac{u ^{\frac{1}{3}}}{2} d u

= 5 \cdot \int_{- 2}^{0} \frac{u ^{\frac{1}{3}}}{2} d u

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \int_{- 2}^{0} u^{\frac{1}{3}} d u

Apply the Power Rule

: [\frac{3}{4} u^{\frac{4}{3}}]_{- 2}^{0}

= 5 \cdot \frac{1}{2} [\frac{3}{4} u^{\frac{4}{3}}]_{- 2}^{0}

Simplify

5 \cdot \frac{1}{2} [\frac{3}{4} u^{\frac{4}{3}}]_{- 2}^{0} : \frac{5}{2} [\frac{3}{4} u^{\frac{4}{3}}]_{- 2}^{0}

= \frac{5}{2} [\frac{3}{4} u^{\frac{4}{3}}]_{- 2}^{0}

Compute the boundaries:

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

= \frac{5}{2} \cdot \frac{3 ( - 1 ) ^{\frac{1}{3}}}{2 ^{\frac{2}{3}}}

Simplify

= \frac{15 ( - 1 ) ^{\frac{1}{3}}}{2 \cdot 2 ^{\frac{2}{3}}}

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What is the integral from-3 to-2 of 5(2x+4)^{1/3} ?
The integral from-3 to-2 of 5(2x+4)^{1/3} is (15(-1)^{1/3})/(2*2^{2/3)}