What is the integral from 1 to 16 of sqrt(3/x) ?

The integral from 1 to 16 of sqrt(3/x) is 6sqrt(3)

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integral from 1 to 16 of sqrt(3/x)

\int_{1}^{16} \frac{3}{x} d x

63

Solution steps

\int_{1}^{16} \frac{3}{x} d x

Apply Integration By Parts

: [\frac{3}{x} x - \int - \frac{3}{2 x ^{\frac{1}{2}}} d x]_{1}^{16}

= [\frac{3}{x} x - \int - \frac{3}{2 x ^{\frac{1}{2}}} d x]_{1}^{16}

\int - \frac{3}{2 x ^{\frac{1}{2}}} d x = - 3 x

= [\frac{3}{x} x - (- 3 x)]_{1}^{16}

Simplify

= [\frac{3}{x} x + 3 x]_{1}^{16}

Compute the boundaries:

63

= 63

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What is the integral from 1 to 16 of sqrt(3/x) ?
The integral from 1 to 16 of sqrt(3/x) is 6sqrt(3)