What is the integral from 4 to 7 of x/(sqrt(x-3)) ?

The integral from 4 to 7 of x/(sqrt(x-3)) is 32/3

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integral from 4 to 7 of x/(sqrt(x-3))

\int_{4}^{7} \frac{x}{x - 3} d x

\frac{32}{3}

Decimal

10.66666 \dots

Solution steps

\int_{4}^{7} \frac{x}{x - 3} d x

Apply u-substitution

: \int_{1}^{2} 2 (u^{2} + 3) d u

= \int_{1}^{2} 2 (u^{2} + 3) d u

Take the constant out:

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

= 2 \cdot \int_{1}^{2} u^{2} + 3 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_{1}^{2} u^{2} d u + \int_{1}^{2} 3 d u)

\int_{1}^{2} u^{2} d u = \frac{7}{3}

\int_{1}^{2} 3 d u = 3

= 2 (\frac{7}{3} + 3)

Simplify

2 (\frac{7}{3} + 3) : \frac{32}{3}

= \frac{32}{3}

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What is the integral from 4 to 7 of x/(sqrt(x-3)) ?
The integral from 4 to 7 of x/(sqrt(x-3)) is 32/3