What is the integral from 6 to 12 of sqrt(3x+2) ?

The integral from 6 to 12 of sqrt(3x+2) is (76sqrt(38)-80sqrt(5))/9

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integral from 6 to 12 of sqrt(3x+2)

\int_{6}^{12} 3 x + 2 d x

\frac{76 38 - 80 5}{9}

Solution steps

\int_{6}^{12} 3 x + 2 d x

Apply u-substitution

: \int_{20}^{38} \frac{u}{3} d u

= \int_{20}^{38} \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_{20}^{38} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{20}^{38}

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{20}^{38}

Compute the boundaries:

\frac{76 38 - 80 5}{3}

= \frac{1}{3} \cdot \frac{76 38 - 80 5}{3}

Simplify

= \frac{76 38 - 80 5}{9}

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What is the integral from 6 to 12 of sqrt(3x+2) ?
The integral from 6 to 12 of sqrt(3x+2) is (76sqrt(38)-80sqrt(5))/9