What is the integral from-4 to 0 of xsqrt(16-x^2) ?

The integral from-4 to 0 of xsqrt(16-x^2) is -64/3

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integral from-4 to 0 of xsqrt(16-x^2)

\int_{- 4}^{0} x 16 - x^{2} d x

- \frac{64}{3}

Decimal

- 21.33333 \dots

Solution steps

\int_{- 4}^{0} x 16 - x^{2} d x

Apply u-substitution

: \int_{0}^{16} - \frac{u}{2} d u

= \int_{0}^{16} - \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}^{16} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{16}

= - \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{16}

Compute the boundaries:

\frac{128}{3}

= - \frac{1}{2} \cdot \frac{128}{3}

Simplify

= - \frac{64}{3}

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What is the integral from-4 to 0 of xsqrt(16-x^2) ?
The integral from-4 to 0 of xsqrt(16-x^2) is -64/3