What is the integral from 1 to 7 of \sqrt[4]{x^3} ?

The integral from 1 to 7 of \sqrt[4]{x^3} is 4*7^{3/4}-4/7

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integral from 1 to 7 of \sqrt[4]{x^3}

\int_{1}^{7} 4 x^{3} d x

4 \cdot 7^{\frac{3}{4}} - \frac{4}{7}

Decimal

16.64263 \dots

Solution steps

\int_{1}^{7} 4 x^{3} d x

4 x^{3} = x^{\frac{3}{4}},

assuming

x \geq 0

= \int_{1}^{7} x^{\frac{3}{4}} d x

Apply the Power Rule

: [\frac{4}{7} x^{\frac{7}{4}}]_{1}^{7}

= [\frac{4}{7} x^{\frac{7}{4}}]_{1}^{7}

Compute the boundaries:

4 \cdot 7^{\frac{3}{4}} - \frac{4}{7}

= 4 \cdot 7^{\frac{3}{4}} - \frac{4}{7}

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What is the integral from 1 to 7 of \sqrt[4]{x^3} ?
The integral from 1 to 7 of \sqrt[4]{x^3} is 4*7^{3/4}-4/7