What is the integral from 0 to 16 of x/(sqrt(4+2x)) ?

The integral from 0 to 16 of x/(sqrt(4+2x)) is 80/3

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integral from 0 to 16 of x/(sqrt(4+2x))

\int_{0}^{16} \frac{x}{4 + 2 x} d x

\frac{80}{3}

Decimal Notation

26.66666 \dots

Solution steps

\int_{0}^{16} \frac{x}{4 + 2 x} d x

Apply u-substitution

: \int_{2}^{6} \frac{u ^{2} - 4}{2} d u

= \int_{2}^{6} \frac{u ^{2} - 4}{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_{2}^{6} u^{2} - 4 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_{2}^{6} u^{2} d u - \int_{2}^{6} 4 d u)

\int_{2}^{6} u^{2} d u = \frac{208}{3}

\int_{2}^{6} 4 d u = 16

= \frac{1}{2} (\frac{208}{3} - 16)

Simplify

\frac{1}{2} (\frac{208}{3} - 16) : \frac{80}{3}

= \frac{80}{3}

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What is the integral from 0 to 16 of x/(sqrt(4+2x)) ?
The integral from 0 to 16 of x/(sqrt(4+2x)) is 80/3