What is the integral from 0 to 16 of x/(sqrt(49+2x)) ?

The integral from 0 to 16 of x/(sqrt(49+2x)) is 46/3

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integral from 0 to 16 of x/(sqrt(49+2x))

\int_{0}^{16} \frac{x}{49 + 2 x} d x

\frac{46}{3}

Decimal

15.33333 \dots

Solution steps

\int_{0}^{16} \frac{x}{49 + 2 x} d x

Apply u-substitution

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

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

Take the constant out:

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

= \frac{1}{2} \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}{2} (\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}{2} (\frac{386}{3} - 98)

Simplify

\frac{1}{2} (\frac{386}{3} - 98) : \frac{46}{3}

= \frac{46}{3}

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What is the integral from 0 to 16 of x/(sqrt(49+2x)) ?
The integral from 0 to 16 of x/(sqrt(49+2x)) is 46/3