What is the integral from 0 to 1 of sqrt(3x+6) ?

The integral from 0 to 1 of sqrt(3x+6) is (18-4sqrt(6))/3

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integral from 0 to 1 of sqrt(3x+6)

\int_{0}^{1} 3 x + 6 d x

\frac{18 - 4 6}{3}

Decimal

2.73401 \dots

Solution steps

\int_{0}^{1} 3 x + 6 d x

Apply u-substitution

: \int_{6}^{9} \frac{u}{3} d u

= \int_{6}^{9} \frac{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_{6}^{9} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{6}^{9}

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{6}^{9}

Compute the boundaries:

18 - 46

= \frac{1}{3} (18 - 46)

Simplify

= \frac{18 - 4 6}{3}

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What is the integral from 0 to 1 of sqrt(3x+6) ?
The integral from 0 to 1 of sqrt(3x+6) is (18-4sqrt(6))/3