What is the integral from 0 to 10 of x/(sqrt(16+2x)) ?

The integral from 0 to 10 of x/(sqrt(16+2x)) is 28/3

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integral from 0 to 10 of x/(sqrt(16+2x))

\int_{0}^{10} \frac{x}{16 + 2 x} d x

\frac{28}{3}

Decimal

9.33333 \dots

Solution steps

\int_{0}^{10} \frac{x}{16 + 2 x} d x

Apply u-substitution

: \int_{4}^{6} \frac{u ^{2} - 16}{2} d u

= \int_{4}^{6} \frac{u ^{2} - 16}{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_{4}^{6} u^{2} - 16 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_{4}^{6} u^{2} d u - \int_{4}^{6} 16 d u)

\int_{4}^{6} u^{2} d u = \frac{152}{3}

\int_{4}^{6} 16 d u = 32

= \frac{1}{2} (\frac{152}{3} - 32)

Simplify

\frac{1}{2} (\frac{152}{3} - 32) : \frac{28}{3}

= \frac{28}{3}

\int_{- 13}^{2} \frac{1}{3 - x} d x

\int_{1}^{2} 10 x^{3} + \frac{6}{x ^{2}} d x

\int_{- 3}^{0} (x^{3} + 2 x^{2} - 3 x) d x

\int_{0}^{x} \frac{1}{θ} e^{\frac{- x}{θ}} d x

\int_{1}^{2} \frac{x}{2 - x ^{2}} d x

What is the integral from 0 to 10 of x/(sqrt(16+2x)) ?
The integral from 0 to 10 of x/(sqrt(16+2x)) is 28/3