What is the integral from 1/3 to 5 of sqrt(1+36x) ?

The integral from 1/3 to 5 of sqrt(1+36x) is (181sqrt(181)-13sqrt(13))/(54)

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integral from 1/3 to 5 of sqrt(1+36x)

\int_{\frac{1}{3}}^{5} 1 + 36 x d x

\frac{181 181 - 13 13}{54}

Decimal

44.22655 \dots

Solution steps

\int_{\frac{1}{3}}^{5} 1 + 36 x d x

Apply u-substitution

= \int_{13}^{181} \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_{13}^{181} u d u

Apply the Power Rule

= \frac{1}{36} [\frac{2}{3} u^{\frac{3}{2}}]_{13}^{181}

Compute the boundaries:

\frac{362 181 - 26 13}{3}

= \frac{1}{36} \cdot \frac{362 181 - 26 13}{3}

Simplify

= \frac{181 181 - 13 13}{54}

What is the integral from 1/3 to 5 of sqrt(1+36x) ?
The integral from 1/3 to 5 of sqrt(1+36x) is (181sqrt(181)-13sqrt(13))/(54)