What is the integral from 2 to 6 of y/(sqrt(y-2)) ?

The integral from 2 to 6 of y/(sqrt(y-2)) is 40/3

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integral from 2 to 6 of y/(sqrt(y-2))

\int_{2}^{6} \frac{y}{y - 2} d y

\frac{40}{3}

Decimal

13.33333 \dots

Solution steps

\int_{2}^{6} \frac{y}{y - 2} d y

Apply u-substitution

: \int_{0}^{2} 2 (u^{2} + 2) d u

= \int_{0}^{2} 2 (u^{2} + 2) d u

Take the constant out:

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

= 2 \cdot \int_{0}^{2} u^{2} + 2 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_{0}^{2} u^{2} d u + \int_{0}^{2} 2 d u)

\int_{0}^{2} u^{2} d u = \frac{8}{3}

\int_{0}^{2} 2 d u = 4

= 2 (\frac{8}{3} + 4)

Simplify

2 (\frac{8}{3} + 4) : \frac{40}{3}

= \frac{40}{3}

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What is the integral from 2 to 6 of y/(sqrt(y-2)) ?
The integral from 2 to 6 of y/(sqrt(y-2)) is 40/3