What is the integral from 0 to 1 of-2xsqrt(9-x^2) ?

The integral from 0 to 1 of-2xsqrt(9-x^2) is -18+(32sqrt(2))/3

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integral from 0 to 1 of-2xsqrt(9-x^2)

\int_{0}^{1} - 2 x 9 - x^{2} d x

- 18 + \frac{32 2}{3}

Solution steps

\int_{0}^{1} - 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}^{1} x 9 - x^{2} d x

Apply u-substitution

: \int_{9}^{8} - \frac{u}{2} d u

= - 2 \cdot \int_{9}^{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}^{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_{8}^{9} u d u))

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{8}^{9}

= - 2 (- (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{8}^{9}))

Simplify

- 2 (- (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{8}^{9})) : - [\frac{2}{3} u^{\frac{3}{2}}]_{8}^{9}

= - [\frac{2}{3} u^{\frac{3}{2}}]_{8}^{9}

Compute the boundaries:

18 - \frac{32 2}{3}

= - (18 - \frac{32 2}{3})

Simplify

= - 18 + \frac{32 2}{3}

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What is the integral from 0 to 1 of-2xsqrt(9-x^2) ?
The integral from 0 to 1 of-2xsqrt(9-x^2) is -18+(32sqrt(2))/3