What is the integral from 0 to 7 of x/(sqrt(36+4x)) ?

The integral from 0 to 7 of x/(sqrt(36+4x)) is 10/3

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integral from 0 to 7 of x/(sqrt(36+4x))

\int_{0}^{7} \frac{x}{36 + 4 x} d x

\frac{10}{3}

Solution steps

\int_{0}^{7} \frac{x}{36 + 4 x} d x

Apply u-substitution

: \int_{6}^{8} \frac{u ^{2} - 36}{8} d u

= \int_{6}^{8} \frac{u ^{2} - 36}{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_{6}^{8} 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{1}{8} (\int_{6}^{8} u^{2} d u - \int_{6}^{8} 36 d u)

\int_{6}^{8} u^{2} d u = \frac{296}{3}

\int_{6}^{8} 36 d u = 72

= \frac{1}{8} (\frac{296}{3} - 72)

Simplify

\frac{1}{8} (\frac{296}{3} - 72) : \frac{10}{3}

= \frac{10}{3}

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What is the integral from 0 to 7 of x/(sqrt(36+4x)) ?
The integral from 0 to 7 of x/(sqrt(36+4x)) is 10/3