What is the integral from 0 to 4.3 of \sqrt[3]{3x+4} ?

The integral from 0 to 4.3 of \sqrt[3]{3x+4} is 9.25492…

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integral from 0 to 4.3 of \sqrt[3]{3x+4}

\int_{0}^{4.3} 3 3 x + 4 d x

9.25492 \dots

Solution steps

\int_{0}^{4.3} 3 3 x + 4 d x

Apply u-substitution

: \int_{4}^{16.9} \frac{3 u}{3} d u

= \int_{4}^{16.9} \frac{3 u}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{4}^{16.9} 3 u d u

Apply the Power Rule

: [\frac{3}{4} u^{\frac{4}{3}}]_{4}^{16.9}

= \frac{1}{3} [\frac{3}{4} u^{\frac{4}{3}}]_{4}^{16.9}

Simplify

= 0.33333 \dots [0.75 u^{\frac{4}{3}}]_{4}^{16.9}

Compute the boundaries:

27.76476 \dots

= 0.33333 \dots \cdot 27.76476 \dots

Simplify

= 9.25492 \dots

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What is the integral from 0 to 4.3 of \sqrt[3]{3x+4} ?
The integral from 0 to 4.3 of \sqrt[3]{3x+4} is 9.25492…