What is the integral from 0 to y of x(y^2-x^2)^{3/2} ?

The integral from 0 to y of x(y^2-x^2)^{3/2} is ((y^2)^{5/2})/5

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integral from 0 to y of x(y^2-x^2)^{3/2}

\int_{0}^{y} x (y^{2} - x^{2})^{\frac{3}{2}} d x

\frac{( y ^{2} ) ^{\frac{5}{2}}}{5}

Solution steps

\int_{0}^{y} x (y^{2} - x^{2})^{\frac{3}{2}} d x

Apply u-substitution

: \int_{y^{2}}^{0} - \frac{u ^{\frac{3}{2}}}{2} d u

= \int_{y^{2}}^{0} - \frac{u ^{\frac{3}{2}}}{2} d u

Take the constant out:

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

= - \frac{1}{2} \cdot \int_{y^{2}}^{0} u^{\frac{3}{2}} d u

Apply the Power Rule

: [\frac{2}{5} u^{\frac{5}{2}}]_{y^{2}}^{0}

= - \frac{1}{2} [\frac{2}{5} u^{\frac{5}{2}}]_{y^{2}}^{0}

Compute the boundaries:

- \frac{2}{5} (y^{2})^{\frac{5}{2}}

= - \frac{1}{2} (- \frac{2}{5} (y^{2})^{\frac{5}{2}})

Simplify

= \frac{( y ^{2} ) ^{\frac{5}{2}}}{5}

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What is the integral from 0 to y of x(y^2-x^2)^{3/2} ?
The integral from 0 to y of x(y^2-x^2)^{3/2} is ((y^2)^{5/2})/5