What is the integral from 0 to 8 of x/(sqrt(49+4x)) ?

The integral from 0 to 8 of x/(sqrt(49+4x)) is 23/6

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integral from 0 to 8 of x/(sqrt(49+4x))

\int_{0}^{8} \frac{x}{49 + 4 x} d x

\frac{23}{6}

Decimal

3.83333 \dots

Solution steps

\int_{0}^{8} \frac{x}{49 + 4 x} d x

Apply u-substitution

: \int_{7}^{9} \frac{u ^{2} - 49}{8} d u

= \int_{7}^{9} \frac{u ^{2} - 49}{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_{7}^{9} u^{2} - 49 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_{7}^{9} u^{2} d u - \int_{7}^{9} 49 d u)

\int_{7}^{9} u^{2} d u = \frac{386}{3}

\int_{7}^{9} 49 d u = 98

= \frac{1}{8} (\frac{386}{3} - 98)

Simplify

\frac{1}{8} (\frac{386}{3} - 98) : \frac{23}{6}

= \frac{23}{6}

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What is the integral from 0 to 8 of x/(sqrt(49+4x)) ?
The integral from 0 to 8 of x/(sqrt(49+4x)) is 23/6