What is the integral from 0 to 8 of sqrt(1+9/4 x) ?

The integral from 0 to 8 of sqrt(1+9/4 x) is (4(38sqrt(19)-2))/(27)

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integral from 0 to 8 of sqrt(1+9/4 x)

\int_{0}^{8} 1 + \frac{9}{4} x d x

\frac{4 ( 38 19 - 2 )}{27}

Decimal

24.24269 \dots

Solution steps

\int_{0}^{8} 1 + \frac{9}{4} x d x

Apply u-substitution

: \int_{1}^{19} \frac{4 u}{9} d u

= \int_{1}^{19} \frac{4 u}{9} d u

Take the constant out:

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

= \frac{4}{9} \cdot \int_{1}^{19} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{19}

= \frac{4}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{19}

Compute the boundaries:

\frac{38 19}{3} - \frac{2}{3}

= \frac{4}{9} (\frac{38 19}{3} - \frac{2}{3})

Simplify

= \frac{4 ( 38 19 - 2 )}{27}

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What is the integral from 0 to 8 of sqrt(1+9/4 x) ?
The integral from 0 to 8 of sqrt(1+9/4 x) is (4(38sqrt(19)-2))/(27)