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

The integral from 1 to 4 of sqrt(3x+1) is (26sqrt(13)-16)/9

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integral from 1 to 4 of sqrt(3x+1)

\int_{1}^{4} 3 x + 1 d x

\frac{26 13 - 16}{9}

Solution steps

\int_{1}^{4} 3 x + 1 d x

Apply u-substitution

: \int_{4}^{13} \frac{u}{3} d u

= \int_{4}^{13} \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_{4}^{13} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{13}

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{13}

Compute the boundaries:

\frac{26 13}{3} - \frac{16}{3}

= \frac{1}{3} (\frac{26 13}{3} - \frac{16}{3})

Simplify

= \frac{26 13 - 16}{9}

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What is the integral from 1 to 4 of sqrt(3x+1) ?
The integral from 1 to 4 of sqrt(3x+1) is (26sqrt(13)-16)/9