What is the integral from 0 to a of 9xsqrt(a^2-x^2) ?

The integral from 0 to a of 9xsqrt(a^2-x^2) is 3(a^2)^{3/2}

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integral from 0 to a of 9xsqrt(a^2-x^2)

\int_{0}^{a} 9 x a^{2} - x^{2} d x

3 (a^{2})^{\frac{3}{2}}

Solution steps

\int_{0}^{a} 9 x a^{2} - x^{2} d x

Take the constant out:

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

= 9 \cdot \int_{0}^{a} x a^{2} - x^{2} d x

Apply u-substitution

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

= 9 \cdot \int_{a^{2}}^{0} - \frac{u}{2} d u

Take the constant out:

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

= 9 (- \frac{1}{2} \cdot \int_{a^{2}}^{0} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{a^{2}}^{0}

= 9 (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{a^{2}}^{0})

Simplify

9 (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{a^{2}}^{0}) : - \frac{9}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{a^{2}}^{0}

= - \frac{9}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{a^{2}}^{0}

Compute the boundaries:

- \frac{2}{3} (a^{2})^{\frac{3}{2}}

= - \frac{9}{2} (- \frac{2}{3} (a^{2})^{\frac{3}{2}})

Simplify

= 3 (a^{2})^{\frac{3}{2}}

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What is the integral from 0 to a of 9xsqrt(a^2-x^2) ?
The integral from 0 to a of 9xsqrt(a^2-x^2) is 3(a^2)^{3/2}