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

The integral from 0 to x of sqrt(1+36t) is ((36x+1)^{3/2}-1)/(54)

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

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

\frac{( 36 x + 1 ) ^{\frac{3}{2}} - 1}{54}

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{2}{3} (1 + 36 x)^{\frac{3}{2}} - \frac{2}{3}

= \frac{1}{36} (\frac{2}{3} (1 + 36 x)^{\frac{3}{2}} - \frac{2}{3})

Simplify

= \frac{( 36 x + 1 ) ^{\frac{3}{2}} - 1}{54}

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What is the integral from 0 to x of sqrt(1+36t) ?
The integral from 0 to x of sqrt(1+36t) is ((36x+1)^{3/2}-1)/(54)