What is the integral from 0 to 2 of 2xsqrt(12-x^2) ?

The integral from 0 to 2 of 2xsqrt(12-x^2) is 16sqrt(3)-(32sqrt(2))/3

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integral from 0 to 2 of 2xsqrt(12-x^2)

\int_{0}^{2} 2 x 12 - x^{2} d x

163 - \frac{32 2}{3}

Decimal Notation

12.62786 \dots

Solution steps

\int_{0}^{2} 2 x 12 - x^{2} d x

Take the constant out:

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

= 2 \cdot \int_{0}^{2} x 12 - x^{2} d x

Apply u-substitution

: \int_{12}^{8} - \frac{u}{2} d u

= 2 \cdot \int_{12}^{8} - \frac{u}{2} d u

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

= 2 (- \int_{8}^{12} - \frac{u}{2} d u)

Take the constant out:

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

= 2 (- (- \frac{1}{2} \cdot \int_{8}^{12} u d u))

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{8}^{12}

= 2 (- (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{8}^{12}))

Simplify

2 (- (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{8}^{12})) : [\frac{2}{3} u^{\frac{3}{2}}]_{8}^{12}

= [\frac{2}{3} u^{\frac{3}{2}}]_{8}^{12}

Compute the boundaries:

163 - \frac{32 2}{3}

= 163 - \frac{32 2}{3}

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What is the integral from 0 to 2 of 2xsqrt(12-x^2) ?
The integral from 0 to 2 of 2xsqrt(12-x^2) is 16sqrt(3)-(32sqrt(2))/3