What is the integral from 0 to 35/36 of sqrt(1+36x) ?

The integral from 0 to 35/36 of sqrt(1+36x) is 215/54

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integral from 0 to 35/36 of sqrt(1+36x)

\int_{0}^{\frac{35}{36}} 1 + 36 x d x

\frac{215}{54}

Solution steps

\int_{0}^{\frac{35}{36}} 1 + 36 x d x

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{430}{3}

= \frac{1}{36} \cdot \frac{430}{3}

Simplify

= \frac{215}{54}

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What is the integral from 0 to 35/36 of sqrt(1+36x) ?
The integral from 0 to 35/36 of sqrt(1+36x) is 215/54