What is the integral from 0 to 2 of sqrt(3x+1) ?

The integral from 0 to 2 of sqrt(3x+1) is (14sqrt(7)-2)/9

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integral from 0 to 2 of sqrt(3x+1)

\int_{0}^{2} 3 x + 1 d x

\frac{14 7 - 2}{9}

Decimal

3.89339 \dots

Solution steps

\int_{0}^{2} 3 x + 1 d x

Apply u-substitution

= \int_{1}^{7} \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}^{7} u d u

Apply the Power Rule

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

Compute the boundaries:

\frac{14 7}{3} - \frac{2}{3}

= \frac{1}{3} (\frac{14 7}{3} - \frac{2}{3})

Simplify

= \frac{14 7 - 2}{9}

What is the integral from 0 to 2 of sqrt(3x+1) ?
The integral from 0 to 2 of sqrt(3x+1) is (14sqrt(7)-2)/9