What is the integral from 0 to 2 of 1/(sqrt(k(2-x))) ?

The integral from 0 to 2 of 1/(sqrt(k(2-x))) is (2sqrt(2))/(sqrt(k))

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integral from 0 to 2 of 1/(sqrt(k(2-x)))

\int_{0}^{2} \frac{1}{k ( 2 - x )} d x

\frac{2 2}{k}

Solution steps

\int_{0}^{2} \frac{1}{k ( 2 - x )} d x

\frac{1}{k ( 2 - x )} = \frac{1}{k} \frac{1}{2 - x}

= \int_{0}^{2} \frac{1}{k} \cdot \frac{1}{2 - x} d x

Take the constant out:

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

= \frac{1}{k} \cdot \int_{0}^{2} \frac{1}{2 - x} d x

Apply u-substitution

: \int_{2}^{0} - \frac{1}{u} d u

= \frac{1}{k} \cdot \int_{2}^{0} - \frac{1}{u} d u

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

= \frac{1}{k} (- \int_{0}^{2} - \frac{1}{u} d u)

Take the constant out:

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

= \frac{1}{k} (- (- \int_{0}^{2} \frac{1}{u} d u))

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{0}^{2}

= \frac{1}{k} (- (- [2 u^{\frac{1}{2}}]_{0}^{2}))

Simplify

\frac{1}{k} (- (- [2 u^{\frac{1}{2}}]_{0}^{2})) : \frac{[ 2 u ] _{0}^{2}}{k}

= \frac{[ 2 u ] _{0}^{2}}{k}

Compute the boundaries:

22

= \frac{2 2}{k}

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What is the integral from 0 to 2 of 1/(sqrt(k(2-x))) ?
The integral from 0 to 2 of 1/(sqrt(k(2-x))) is (2sqrt(2))/(sqrt(k))