What is the integral from 1 to 9 of 1/(8sqrt(uv)) ?

The integral from 1 to 9 of 1/(8sqrt(uv)) is 1/(2sqrt(v))

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integral from 1 to 9 of 1/(8sqrt(uv))

\int_{1}^{9} \frac{1}{8 uv} d u

\frac{1}{2 v}

Solution steps

\int_{1}^{9} \frac{1}{8 uv} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int_{1}^{9} \frac{1}{vu} d u

\frac{1}{vu} = \frac{1}{v} \frac{1}{u}

= \frac{1}{8} \cdot \int_{1}^{9} \frac{1}{v} \cdot \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}{8} \cdot \frac{1}{v} \cdot \int_{1}^{9} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{1}^{9}

= \frac{1}{8} \cdot \frac{1}{v} [2 u^{\frac{1}{2}}]_{1}^{9}

Simplify

\frac{1}{8} \cdot \frac{1}{v} [2 u^{\frac{1}{2}}]_{1}^{9} : \frac{[ 2 u ] _{1}^{9}}{8 v}

= \frac{[ 2 u ] _{1}^{9}}{8 v}

Compute the boundaries:

4

= \frac{4}{8 v}

Simplify

= \frac{1}{2 v}

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What is the integral from 1 to 9 of 1/(8sqrt(uv)) ?
The integral from 1 to 9 of 1/(8sqrt(uv)) is 1/(2sqrt(v))