What is the integral from 1 to 4 of sqrt(3x-2) ?

The integral from 1 to 4 of sqrt(3x-2) is (20sqrt(10)-2)/9

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integral from 1 to 4 of sqrt(3x-2)

\int_{1}^{4} 3 x - 2 d x

\frac{20 10 - 2}{9}

Decimal Notation

6.80506 \dots

Solution steps

\int_{1}^{4} 3 x - 2 d x

Apply u-substitution

: \int_{1}^{10} \frac{u}{3} d u

= \int_{1}^{10} \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_{1}^{10} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{10}

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{10}

Compute the boundaries:

\frac{20 10}{3} - \frac{2}{3}

= \frac{1}{3} (\frac{20 10}{3} - \frac{2}{3})

Simplify

= \frac{20 10 - 2}{9}

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What is the integral from 1 to 4 of sqrt(3x-2) ?
The integral from 1 to 4 of sqrt(3x-2) is (20sqrt(10)-2)/9