What is the integral from 0 to 4 of x/(sqrt(16+5x)) ?

The integral from 0 to 4 of x/(sqrt(16+5x)) is 112/75

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integral from 0 to 4 of x/(sqrt(16+5x))

\int_{0}^{4} \frac{x}{16 + 5 x} d x

\frac{112}{75}

Solution steps

\int_{0}^{4} \frac{x}{16 + 5 x} d x

Apply u-substitution

: \int_{4}^{6} \frac{2 ( u ^{2} - 16 )}{25} d u

= \int_{4}^{6} \frac{2 ( u ^{2} - 16 )}{25} d u

Take the constant out:

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

= \frac{2}{25} \cdot \int_{4}^{6} 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{2}{25} (\int_{4}^{6} u^{2} d u - \int_{4}^{6} 16 d u)

\int_{4}^{6} u^{2} d u = \frac{152}{3}

\int_{4}^{6} 16 d u = 32

= \frac{2}{25} (\frac{152}{3} - 32)

Simplify

\frac{2}{25} (\frac{152}{3} - 32) : \frac{112}{75}

= \frac{112}{75}

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What is the integral from 0 to 4 of x/(sqrt(16+5x)) ?
The integral from 0 to 4 of x/(sqrt(16+5x)) is 112/75