What is the integral from 0 to 4 of \sqrt[3]{2x} ?

The integral from 0 to 4 of \sqrt[3]{2x} is 6

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integral from 0 to 4 of \sqrt[3]{2x}

\int_{0}^{4} 3 2 x d x

6

Solution steps

\int_{0}^{4} 3 2 x d x

Apply Integration By Parts

: [32 x 3 x - \int \frac{2 ^{\frac{1}{3}}}{3} x^{\frac{1}{3}} d x]_{0}^{4}

= [32 x 3 x - \int \frac{2 ^{\frac{1}{3}}}{3} x^{\frac{1}{3}} d x]_{0}^{4}

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

= [32 x 3 x - \frac{1}{2 \cdot 2 ^{\frac{2}{3}}} x^{\frac{4}{3}}]_{0}^{4}

Compute the boundaries:

6

= 6

\frac{x ^{2} - 4 x + 3}{x ^{2}}

7 y^{^{''}} + 7 y^{^{'}} - 14 y = 0

7 (3 x^{2} ln (5 x) + x^{2})

x \to 2 + lim (8 - x^{3})

\frac{d}{d t} (3 t cos (t))

What is the integral from 0 to 4 of \sqrt[3]{2x} ?
The integral from 0 to 4 of \sqrt[3]{2x} is 6