What is the integral from 1/2 to 1 of 7/(\sqrt[3]{2x-1)} ?

The integral from 1/2 to 1 of 7/(\sqrt[3]{2x-1)} is 21/4

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integral from 1/2 to 1 of 7/(\sqrt[3]{2x-1)}

\frac{21}{4}

Decimal Notation

5.25

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}{2} u^{\frac{2}{3}}]_{0}^{1}

= 7 \cdot \frac{1}{2} [\frac{3}{2} u^{\frac{2}{3}}]_{0}^{1}

Simplify

7 \cdot \frac{1}{2} [\frac{3}{2} u^{\frac{2}{3}}]_{0}^{1} : \frac{7}{2} [\frac{3}{2} u^{\frac{2}{3}}]_{0}^{1}

= \frac{7}{2} [\frac{3}{2} u^{\frac{2}{3}}]_{0}^{1}

Compute the boundaries:

\frac{3}{2}

= \frac{7}{2} \cdot \frac{3}{2}

Simplify

= \frac{21}{4}

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What is the integral from 1/2 to 1 of 7/(\sqrt[3]{2x-1)} ?
The integral from 1/2 to 1 of 7/(\sqrt[3]{2x-1)} is 21/4