What is the integral from 0 to 2 of sqrt(9x+19) ?

The integral from 0 to 2 of sqrt(9x+19) is (74sqrt(37)-38sqrt(19))/(27)

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integral from 0 to 2 of sqrt(9x+19)

\int_{0}^{2} 9 x + 19 d x

\frac{74 37 - 38 19}{27}

Decimal Notation

10.53652 \dots

Solution steps

\int_{0}^{2} 9 x + 19 d x

Apply u-substitution

: \int_{19}^{37} \frac{u}{9} d u

= \int_{19}^{37} \frac{u}{9} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{9} \cdot \int_{19}^{37} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{19}^{37}

= \frac{1}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{19}^{37}

Compute the boundaries:

\frac{74 37 - 38 19}{3}

= \frac{1}{9} \cdot \frac{74 37 - 38 19}{3}

Simplify

= \frac{74 37 - 38 19}{27}

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What is the integral from 0 to 2 of sqrt(9x+19) ?
The integral from 0 to 2 of sqrt(9x+19) is (74sqrt(37)-38sqrt(19))/(27)