What is the integral from 0 to 4 of sqrt(1+36x) ?

The integral from 0 to 4 of sqrt(1+36x) is (145sqrt(145)-1)/(54)

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integral from 0 to 4 of sqrt(1+36x)

\int_{0}^{4} 1 + 36 x d x

\frac{145 145 - 1}{54}

Decimal

32.31539 \dots

Solution steps

\int_{0}^{4} 1 + 36 x d x

Apply u-substitution

= \int_{1}^{145} \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}^{145} u d u

Apply the Power Rule

= \frac{1}{36} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{145}

Compute the boundaries:

\frac{290 145}{3} - \frac{2}{3}

= \frac{1}{36} (\frac{290 145}{3} - \frac{2}{3})

Simplify

= \frac{145 145 - 1}{54}

What is the integral from 0 to 4 of sqrt(1+36x) ?
The integral from 0 to 4 of sqrt(1+36x) is (145sqrt(145)-1)/(54)