What is the integral from 1 to 5 of sqrt(9x-4) ?

The integral from 1 to 5 of sqrt(9x-4) is (82sqrt(41)-10sqrt(5))/(27)

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integral from 1 to 5 of sqrt(9x-4)

\int_{1}^{5} 9 x - 4 d x

\frac{82 41 - 10 5}{27}

Decimal

18.61835 \dots

Solution steps

\int_{1}^{5} 9 x - 4 d x

Apply u-substitution

: \int_{5}^{41} \frac{u}{9} d u

= \int_{5}^{41} \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_{5}^{41} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{5}^{41}

= \frac{1}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{5}^{41}

Compute the boundaries:

\frac{82 41 - 10 5}{3}

= \frac{1}{9} \cdot \frac{82 41 - 10 5}{3}

Simplify

= \frac{82 41 - 10 5}{27}

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What is the integral from 1 to 5 of sqrt(9x-4) ?
The integral from 1 to 5 of sqrt(9x-4) is (82sqrt(41)-10sqrt(5))/(27)