What is the integral from 0 to 15 of x/(sqrt(4+3x)) ?

The integral from 0 to 15 of x/(sqrt(4+3x)) is 550/27

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integral from 0 to 15 of x/(sqrt(4+3x))

\int_{0}^{15} \frac{x}{4 + 3 x} d x

\frac{550}{27}

Decimal

20.37037 \dots

Solution steps

\int_{0}^{15} \frac{x}{4 + 3 x} d x

Apply u-substitution

: \int_{2}^{7} \frac{2 ( u ^{2} - 4 )}{9} d u

= \int_{2}^{7} \frac{2 ( u ^{2} - 4 )}{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_{2}^{7} 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{2}{9} (\int_{2}^{7} u^{2} d u - \int_{2}^{7} 4 d u)

\int_{2}^{7} u^{2} d u = \frac{335}{3}

\int_{2}^{7} 4 d u = 20

= \frac{2}{9} (\frac{335}{3} - 20)

Simplify

\frac{2}{9} (\frac{335}{3} - 20) : \frac{550}{27}

= \frac{550}{27}

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What is the integral from 0 to 15 of x/(sqrt(4+3x)) ?
The integral from 0 to 15 of x/(sqrt(4+3x)) is 550/27