What is the integral from 2 to 4 of (sqrt(x-2))/x ?

The integral from 2 to 4 of (sqrt(x-2))/x is 2sqrt(2)(-pi/4+1)

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integral from 2 to 4 of (sqrt(x-2))/x

\int_{2}^{4} \frac{x - 2}{x} d x

22 (- \frac{π}{4} + 1)

Decimal Notation

0.60698 \dots

Solution steps

\int_{2}^{4} \frac{x - 2}{x} d x

Apply u-substitution

: \int_{0}^{2} \frac{2 u ^{2}}{u ^{2} + 2} d u

= \int_{0}^{2} \frac{2 u ^{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} \frac{u ^{2}}{u ^{2} + 2} d u

Apply Integral Substitution

= 2 \cdot \int_{0}^{1} \frac{2 v ^{2}}{v ^{2} + 1} d v

Take the constant out:

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

= 22 \cdot \int_{0}^{1} \frac{v ^{2}}{v ^{2} + 1} d v

\frac{v ^{2}}{v ^{2} + 1} = - \frac{1}{v ^{2} + 1} + 1

= 22 \cdot \int_{0}^{1} - \frac{1}{v ^{2} + 1} + 1 d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 22 (- \int_{0}^{1} \frac{1}{v ^{2} + 1} d v + \int_{0}^{1} 1 d v)

\int_{0}^{1} \frac{1}{v ^{2} + 1} d v = \frac{π}{4}

\int_{0}^{1} 1 d v = 1

= 22 (- \frac{π}{4} + 1)

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What is the integral from 2 to 4 of (sqrt(x-2))/x ?
The integral from 2 to 4 of (sqrt(x-2))/x is 2sqrt(2)(-pi/4+1)