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

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

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

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

- y (- 2 + 2 y)

Solution steps

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

Apply u-substitution

: \int_{\frac{1}{y}}^{1} - \frac{y}{u ^{\frac{3}{2}}} d u

= \int_{\frac{1}{y}}^{1} - \frac{y}{u ^{\frac{3}{2}}} d u

Take the constant out:

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

= - y \cdot \int_{\frac{1}{y}}^{1} \frac{1}{u ^{\frac{3}{2}}} d u

Apply the Power Rule

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

= - y [- \frac{2}{u}]_{\frac{1}{y}}^{1}

Compute the boundaries:

- 2 + 2 y

= - y (- 2 + 2 y)

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