What is the integral from 0 to 15 of x/(sqrt(36+3x)) ?

The integral from 0 to 15 of x/(sqrt(36+3x)) is 14

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integral from 0 to 15 of x/(sqrt(36+3x))

\int_{0}^{15} \frac{x}{36 + 3 x} d x

14

Solution steps

\int_{0}^{15} \frac{x}{36 + 3 x} d x

Apply u-substitution

: \int_{6}^{9} \frac{2 ( u ^{2} - 36 )}{9} d u

= \int_{6}^{9} \frac{2 ( u ^{2} - 36 )}{9} d u

Take the constant out:

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

= \frac{2}{9} \cdot \int_{6}^{9} u^{2} - 36 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}{9} (\int_{6}^{9} u^{2} d u - \int_{6}^{9} 36 d u)

\int_{6}^{9} u^{2} d u = 171

\int_{6}^{9} 36 d u = 108

= \frac{2}{9} (171 - 108)

Simplify

\frac{2}{9} (171 - 108) : 14

= 14

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What is the integral from 0 to 15 of x/(sqrt(36+3x)) ?
The integral from 0 to 15 of x/(sqrt(36+3x)) is 14