What is the integral from 0 to 12 of x/(sqrt(16+4x)) ?

The integral from 0 to 12 of x/(sqrt(16+4x)) is 32/3

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integral from 0 to 12 of x/(sqrt(16+4x))

\int_{0}^{12} \frac{x}{16 + 4 x} d x

\frac{32}{3}

Decimal

10.66666 \dots

Solution steps

\int_{0}^{12} \frac{x}{16 + 4 x} d x

Apply u-substitution

: \int_{4}^{8} \frac{u ^{2} - 16}{8} d u

= \int_{4}^{8} \frac{u ^{2} - 16}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int_{4}^{8} u^{2} - 16 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}{8} (\int_{4}^{8} u^{2} d u - \int_{4}^{8} 16 d u)

\int_{4}^{8} u^{2} d u = \frac{448}{3}

\int_{4}^{8} 16 d u = 64

= \frac{1}{8} (\frac{448}{3} - 64)

Simplify

\frac{1}{8} (\frac{448}{3} - 64) : \frac{32}{3}

= \frac{32}{3}

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What is the integral from 0 to 12 of x/(sqrt(16+4x)) ?
The integral from 0 to 12 of x/(sqrt(16+4x)) is 32/3