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

The integral from 1 to 6 of sqrt(6x+1) is (37sqrt(37)-7sqrt(7))/9

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

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

\frac{37 37 - 7 7}{9}

Decimal Notation

22.94910 \dots

Solution steps

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

Apply u-substitution

: \int_{7}^{37} \frac{u}{6} d u

= \int_{7}^{37} \frac{u}{6} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{6} \cdot \int_{7}^{37} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{7}^{37}

= \frac{1}{6} [\frac{2}{3} u^{\frac{3}{2}}]_{7}^{37}

Compute the boundaries:

\frac{74 37 - 14 7}{3}

= \frac{1}{6} \cdot \frac{74 37 - 14 7}{3}

Simplify

= \frac{37 37 - 7 7}{9}

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What is the integral from 1 to 6 of sqrt(6x+1) ?
The integral from 1 to 6 of sqrt(6x+1) is (37sqrt(37)-7sqrt(7))/9