What is the integral from 0 to 1 of sqrt(x/y) ?

The integral from 0 to 1 of sqrt(x/y) is sqrt(1/y)-1/(3sqrt(y))

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integral from 0 to 1 of sqrt(x/y)

\int_{0}^{1} \frac{x}{y} d x

\frac{1}{y} - \frac{1}{3 y}

Solution steps

\int_{0}^{1} \frac{x}{y} d x

Apply Integration By Parts

: [\frac{x}{y} x - \int \frac{x}{2 y} d x]_{0}^{1}

= [\frac{x}{y} x - \int \frac{x}{2 y} d x]_{0}^{1}

\int \frac{x}{2 y} d x = \frac{x ^{\frac{3}{2}}}{3 y}

= [\frac{x}{y} x - \frac{x ^{\frac{3}{2}}}{3 y}]_{0}^{1}

Compute the boundaries:

\frac{1}{y} - \frac{1}{3 y}

= \frac{1}{y} - \frac{1}{3 y}

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\int_{0}^{1} (2 x^{2} + 3 x - 4) d x

What is the integral from 0 to 1 of sqrt(x/y) ?
The integral from 0 to 1 of sqrt(x/y) is sqrt(1/y)-1/(3sqrt(y))