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

The integral from 0 to 4 of x/(sqrt(8-x)) is 2(-(16sqrt(2)-8)/3+16sqrt(2)-16)

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

\int_{0}^{4} \frac{x}{8 - x} d x

2 (- \frac{16 2 - 8}{3} + 162 - 16)

Decimal Notation

3.50322 \dots

Solution steps

\int_{0}^{4} \frac{x}{8 - x} d x

Apply u-substitution

: \int_{22}^{2} - 2 (- u^{2} + 8) d u

= \int_{22}^{2} - 2 (- u^{2} + 8) d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{2}^{22} - 2 (- u^{2} + 8) d u

Take the constant out:

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

= - (- 2 \cdot \int_{2}^{22} - u^{2} + 8 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

= - (- 2 (- \int_{2}^{22} u^{2} d u + \int_{2}^{22} 8 d u))

\int_{2}^{22} u^{2} d u = \frac{16 2 - 8}{3}

\int_{2}^{22} 8 d u = 162 - 16

= - (- 2 (- \frac{16 2 - 8}{3} + 162 - 16))

Simplify

= 2 (- \frac{16 2 - 8}{3} + 162 - 16)

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What is the integral from 0 to 4 of x/(sqrt(8-x)) ?
The integral from 0 to 4 of x/(sqrt(8-x)) is 2(-(16sqrt(2)-8)/3+16sqrt(2)-16)