What is the integral from 1 to 3 of 2\sqrt[4]{x^3} ?

The integral from 1 to 3 of 2\sqrt[4]{x^3} is (2(12*3^{3/4}-4))/7

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integral from 1 to 3 of 2\sqrt[4]{x^3}

\frac{2 ( 12 \cdot 3 ^{\frac{3}{4}} - 4 )}{7}

Decimal

6.67259 \dots

Solution steps

Take the constant out:

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

assuming

x \geq 0

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

Apply the Power Rule

= 2 [\frac{4}{7} x^{\frac{7}{4}}]_{1}^{3}

Compute the boundaries:

\frac{12 \cdot 3 ^{\frac{3}{4}}}{7} - \frac{4}{7}

= 2 (\frac{12 \cdot 3 ^{\frac{3}{4}}}{7} - \frac{4}{7})

Simplify

= \frac{2 ( 12 \cdot 3 ^{\frac{3}{4}} - 4 )}{7}

What is the integral from 1 to 3 of 2\sqrt[4]{x^3} ?
The integral from 1 to 3 of 2\sqrt[4]{x^3} is (2(12*3^{3/4}-4))/7