What is the integral from 0 to 1 of sqrt(1+12x) ?

The integral from 0 to 1 of sqrt(1+12x) is (13sqrt(13)-1)/(18)

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integral from 0 to 1 of sqrt(1+12x)

\int_{0}^{1} 1 + 12 x d x

\frac{13 13 - 1}{18}

Solution steps

\int_{0}^{1} 1 + 12 x d x

Apply u-substitution

: \int_{1}^{13} \frac{u}{12} d u

= \int_{1}^{13} \frac{u}{12} d u

Take the constant out:

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

= \frac{1}{12} \cdot \int_{1}^{13} u d u

Apply the Power Rule

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

= \frac{1}{12} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{13}

Compute the boundaries:

\frac{26 13}{3} - \frac{2}{3}

= \frac{1}{12} (\frac{26 13}{3} - \frac{2}{3})

Simplify

= \frac{13 13 - 1}{18}

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What is the integral from 0 to 1 of sqrt(1+12x) ?
The integral from 0 to 1 of sqrt(1+12x) is (13sqrt(13)-1)/(18)