What is the integral from 0 to 8 of x/(sqrt(9+2x)) ?

The integral from 0 to 8 of x/(sqrt(9+2x)) is 22/3

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integral from 0 to 8 of x/(sqrt(9+2x))

\int_{0}^{8} \frac{x}{9 + 2 x} d x

\frac{22}{3}

Decimal

7.33333 \dots

Solution steps

\int_{0}^{8} \frac{x}{9 + 2 x} d x

Apply u-substitution

: \int_{3}^{5} \frac{u ^{2} - 9}{2} d u

= \int_{3}^{5} \frac{u ^{2} - 9}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{3}^{5} u^{2} - 9 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{3}^{5} u^{2} d u - \int_{3}^{5} 9 d u)

\int_{3}^{5} u^{2} d u = \frac{98}{3}

\int_{3}^{5} 9 d u = 18

= \frac{1}{2} (\frac{98}{3} - 18)

Simplify

\frac{1}{2} (\frac{98}{3} - 18) : \frac{22}{3}

= \frac{22}{3}

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What is the integral from 0 to 8 of x/(sqrt(9+2x)) ?
The integral from 0 to 8 of x/(sqrt(9+2x)) is 22/3