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

The integral from 1 to 4 of sqrt(1+36x) is (145sqrt(145)-37sqrt(37))/(54)

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

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

\frac{145 145 - 37 37}{54}

Decimal Notation

28.16609 \dots

Solution steps

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

Apply u-substitution

: \int_{37}^{145} \frac{u}{36} d u

= \int_{37}^{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_{37}^{145} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{37}^{145}

= \frac{1}{36} [\frac{2}{3} u^{\frac{3}{2}}]_{37}^{145}

Compute the boundaries:

\frac{290 145 - 74 37}{3}

= \frac{1}{36} \cdot \frac{290 145 - 74 37}{3}

Simplify

= \frac{145 145 - 37 37}{54}

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What is the integral from 1 to 4 of sqrt(1+36x) ?
The integral from 1 to 4 of sqrt(1+36x) is (145sqrt(145)-37sqrt(37))/(54)