What is the integral from 0 to 8 of xsqrt(64-x^2) ?

The integral from 0 to 8 of xsqrt(64-x^2) is 512/3

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integral from 0 to 8 of xsqrt(64-x^2)

\int_{0}^{8} x 64 - x^{2} d x

\frac{512}{3}

Decimal

170.66666 \dots

Solution steps

\int_{0}^{8} x 64 - x^{2} d x

Apply u-substitution

= \int_{64}^{0} - \frac{u}{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{64} - \frac{u}{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_{0}^{64} u d u)

Apply the Power Rule

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{64})

Simplify

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{64}

Compute the boundaries:

\frac{1024}{3}

= \frac{1}{2} \cdot \frac{1024}{3}

Simplify

= \frac{512}{3}

What is the integral from 0 to 8 of xsqrt(64-x^2) ?
The integral from 0 to 8 of xsqrt(64-x^2) is 512/3