What is the integral from 0 to 1 of sqrt(1+36x) ?

The integral from 0 to 1 of sqrt(1+36x) is (37sqrt(37)-1)/(54)

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integral from 0 to 1 of sqrt(1+36x)

\int_{0}^{1} 1 + 36 x d x

\frac{37 37 - 1}{54}

Decimal Notation

4.14930 \dots

Solution steps

\int_{0}^{1} 1 + 36 x d x

Apply u-substitution

: \int_{1}^{37} \frac{u}{36} d u

= \int_{1}^{37} \frac{u}{36} d u

Take the constant out:

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

= \frac{1}{36} \cdot \int_{1}^{37} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{37}

= \frac{1}{36} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{37}

Compute the boundaries:

\frac{74 37}{3} - \frac{2}{3}

= \frac{1}{36} (\frac{74 37}{3} - \frac{2}{3})

Simplify

= \frac{37 37 - 1}{54}

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What is the integral from 0 to 1 of sqrt(1+36x) ?
The integral from 0 to 1 of sqrt(1+36x) is (37sqrt(37)-1)/(54)