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

The integral from 2 to 4 of sqrt(16-x^2)x is 8sqrt(3)

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

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

83

Decimal

13.85640 \dots

Solution steps

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

Apply u-substitution

: \int_{12}^{0} - \frac{u}{2} d u

= \int_{12}^{0} - \frac{u}{2} d u

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

= - \int_{0}^{12} - \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}^{12} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{12}

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{12})

Simplify

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{12}

Compute the boundaries:

163

= \frac{1}{2} \cdot 163

Simplify

= 83

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What is the integral from 2 to 4 of sqrt(16-x^2)x ?
The integral from 2 to 4 of sqrt(16-x^2)x is 8sqrt(3)