What is the integral from 1 to 2 of 16\sqrt[3]{2x-3} ?

The integral from 1 to 2 of 16\sqrt[3]{2x-3} is 0

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integral from 1 to 2 of 16\sqrt[3]{2x-3}

0

Solution steps

Take the constant out:

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

Apply u-substitution

Take the constant out:

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

Apply the Power Rule

: [\frac{3}{4} u^{\frac{4}{3}}]_{- 1}^{1}

= 16 \cdot \frac{1}{2} [\frac{3}{4} u^{\frac{4}{3}}]_{- 1}^{1}

Simplify

16 \cdot \frac{1}{2} [\frac{3}{4} u^{\frac{4}{3}}]_{- 1}^{1} : 8 [\frac{3}{4} u^{\frac{4}{3}}]_{- 1}^{1}

= 8 [\frac{3}{4} u^{\frac{4}{3}}]_{- 1}^{1}

Compute the boundaries:

0

= 8 \cdot 0

Simplify

= 0

\frac{\partial}{\partial x} (x sin (v))

\frac{d}{d x} (cosh^{2} (x) - sinh^{2} (x))

f (x) = x^{5} - x^{3} + 3

(2 x + 1) ln (x)

What is the integral from 1 to 2 of 16\sqrt[3]{2x-3} ?
The integral from 1 to 2 of 16\sqrt[3]{2x-3} is 0