What is the integral from y^2 to y of sqrt(x/y) ?

The integral from y^2 to y of sqrt(x/y) is y+(-y+y^{5/2})/3-y^2sqrt(y)

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integral from y^2 to y of sqrt(x/y)

\int_{y^{2}}^{y} \frac{x}{y} d x

y + \frac{- y + y ^{\frac{5}{2}}}{3} - y^{2} y

Solution steps

\int_{y^{2}}^{y} \frac{x}{y} d x

Apply Integration By Parts

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

\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}]_{y^{2}}^{y}

Compute the boundaries:

y - \frac{y}{3} - y^{2} y + \frac{y ^{\frac{5}{2}}}{3}

= y - \frac{y}{3} - y^{2} y + \frac{y ^{\frac{5}{2}}}{3}

Simplify

= y + \frac{- y + y ^{\frac{5}{2}}}{3} - y^{2} y

What is the integral from y^2 to y of sqrt(x/y) ?
The integral from y^2 to y of sqrt(x/y) is y+(-y+y^{5/2})/3-y^2sqrt(y)