What is the integral from-3 to 0 of 4x^{1/3} ?

The integral from-3 to 0 of 4x^{1/3} is 9(-1)^{1/3}*3^{1/3}

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integral from-3 to 0 of 4x^{1/3}

\int_{- 3}^{0} 4 x^{\frac{1}{3}} d x

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

Solution steps

\int_{- 3}^{0} 4 x^{\frac{1}{3}} d x

Take the constant out:

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

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

Apply the Power Rule

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

= 4 [\frac{3}{4} x^{\frac{4}{3}}]_{- 3}^{0}

Compute the boundaries:

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

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

Simplify

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

\int x^{3} 25 - x^{2} d x

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

\frac{d}{d x} (1 - ∣ x - 1 ∣)

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

What is the integral from-3 to 0 of 4x^{1/3} ?
The integral from-3 to 0 of 4x^{1/3} is 9(-1)^{1/3}*3^{1/3}