What is the integral from 0 to 1 of (1+7x)^{1/3} ?

The integral from 0 to 1 of (1+7x)^{1/3} is 45/28

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integral from 0 to 1 of (1+7x)^{1/3}

\int_{0}^{1} (1 + 7 x)^{\frac{1}{3}} d x

\frac{45}{28}

Decimal Notation

1.60714 \dots

Solution steps

\int_{0}^{1} (1 + 7 x)^{\frac{1}{3}} d x

Apply u-substitution

: \int_{1}^{8} \frac{u ^{\frac{1}{3}}}{7} d u

= \int_{1}^{8} \frac{u ^{\frac{1}{3}}}{7} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int_{1}^{8} u^{\frac{1}{3}} d u

Apply the Power Rule

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

= \frac{1}{7} [\frac{3}{4} u^{\frac{4}{3}}]_{1}^{8}

Compute the boundaries:

\frac{45}{4}

= \frac{1}{7} \cdot \frac{45}{4}

Simplify

= \frac{45}{28}

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What is the integral from 0 to 1 of (1+7x)^{1/3} ?
The integral from 0 to 1 of (1+7x)^{1/3} is 45/28