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

The integral from 0 to 4 of 4xsqrt(16-x^2) is 256/3

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

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

\frac{256}{3}

Decimal

85.33333 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 4 \cdot \int_{16}^{0} - \frac{u}{2} d u

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

= 4 (- \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

= 4 (- (- \frac{1}{2} \cdot \int_{0}^{16} u d u))

Apply the Power Rule

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

= 4 (- (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{16}))

Simplify

4 (- (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{16})) : 2 [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{16}

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

Compute the boundaries:

\frac{128}{3}

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

Simplify

= \frac{256}{3}

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What is the integral from 0 to 4 of 4xsqrt(16-x^2) ?
The integral from 0 to 4 of 4xsqrt(16-x^2) is 256/3