What is the integral from 0 to 20 of x/(sqrt(9+2x)) ?

The integral from 0 to 20 of x/(sqrt(9+2x)) is 104/3

Popular >

integral from 0 to 20 of x/(sqrt(9+2x))

\int_{0}^{20} \frac{x}{9 + 2 x} d x

\frac{104}{3}

Decimal

34.66666 \dots

Solution steps

\int_{0}^{20} \frac{x}{9 + 2 x} d x

Apply u-substitution

: \int_{3}^{7} \frac{u ^{2} - 9}{2} d u

= \int_{3}^{7} \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}^{7} 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}^{7} u^{2} d u - \int_{3}^{7} 9 d u)

\int_{3}^{7} u^{2} d u = \frac{316}{3}

\int_{3}^{7} 9 d u = 36

= \frac{1}{2} (\frac{316}{3} - 36)

Simplify

\frac{1}{2} (\frac{316}{3} - 36) : \frac{104}{3}

= \frac{104}{3}

\int_{π}^{2 π} cos^{2} (θ) d θ

\int_{0}^{1} 2 π x (6 x (x - 1)^{2}) d x

\int_{- 5}^{5} \frac{2}{x ^{2} + 2} d x

\int_{- 3}^{- 1} \frac{2}{r ^{3}} d r

\int_{y^{2}}^{y} 1 d x

What is the integral from 0 to 20 of x/(sqrt(9+2x)) ?
The integral from 0 to 20 of x/(sqrt(9+2x)) is 104/3