What is the integral from 0 to 3 of 2xsqrt(9-x^2) ?

The integral from 0 to 3 of 2xsqrt(9-x^2) is 18

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integral from 0 to 3 of 2xsqrt(9-x^2)

\int_{0}^{3} 2 x 9 - x^{2} d x

18

Solution steps

\int_{0}^{3} 2 x 9 - 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}^{3} x 9 - x^{2} d x

Apply u-substitution

: \int_{9}^{0} - \frac{u}{2} d u

= 2 \cdot \int_{9}^{0} - \frac{u}{2} d u

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

= 2 (- \int_{0}^{9} - \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_{0}^{9} u d u))

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{9}

= 2 (- (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{9}))

Simplify

2 (- (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{9})) : [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{9}

= [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{9}

Compute the boundaries:

18

= 18

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What is the integral from 0 to 3 of 2xsqrt(9-x^2) ?
The integral from 0 to 3 of 2xsqrt(9-x^2) is 18