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

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

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

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

- x (- 2 + 2 x)

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Compute the boundaries:

- 2 + 2 x

= - x (- 2 + 2 x)

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