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

The integral from-2 to 2 of (x-3)sqrt(4-x^2) is -6pi

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integral from-2 to 2 of (x-3)sqrt(4-x^2)

\int_{- 2}^{2} (x - 3) 4 - x^{2} d x

- 6 π

Decimal

- 18.84955 \dots

Solution steps

\int_{- 2}^{2} (x - 3) 4 - x^{2} d x

Apply Integration By Parts

: [(x - 3) (2 arcsin (\frac{1}{2} x) + sin (2 arcsin (\frac{1}{2} x))) - \int 2 arcsin (\frac{1}{2} x) + sin (2 arcsin (\frac{1}{2} x)) d x]_{- 2}^{2}

= [(x - 3) (2 arcsin (\frac{1}{2} x) + sin (2 arcsin (\frac{1}{2} x))) - \int 2 arcsin (\frac{1}{2} x) + sin (2 arcsin (\frac{1}{2} x)) d x]_{- 2}^{2}

\int 2 arcsin (\frac{1}{2} x) + sin (2 arcsin (\frac{1}{2} x)) d x = x (\frac{1}{2} x 4 - x^{2} + 2 arcsin (\frac{x}{2})) + \frac{1}{3} (- x^{2} + 4)^{\frac{3}{2}}

= [(x - 3) (2 arcsin (\frac{1}{2} x) + sin (2 arcsin (\frac{1}{2} x))) - (x (\frac{1}{2} x 4 - x^{2} + 2 arcsin (\frac{x}{2})) + \frac{1}{3} (- x^{2} + 4)^{\frac{3}{2}})]_{- 2}^{2}

Simplify

[(x - 3) (2 arcsin (\frac{1}{2} x) + sin (2 arcsin (\frac{1}{2} x))) - (x (\frac{1}{2} x 4 - x^{2} + 2 arcsin (\frac{x}{2})) + \frac{1}{3} (- x^{2} + 4)^{\frac{3}{2}})]_{- 2}^{2} : [\frac{1}{6} (- 3 x^{2} - x^{2} + 4 + 6 x sin (2 arcsin (\frac{x}{2})) - 2 (- x^{2} + 4)^{\frac{3}{2}} - 36 arcsin (\frac{x}{2}) - 18 sin (2 arcsin (\frac{x}{2})))]_{- 2}^{2}

= [\frac{1}{6} (- 3 x^{2} - x^{2} + 4 + 6 x sin (2 arcsin (\frac{x}{2})) - 2 (- x^{2} + 4)^{\frac{3}{2}} - 36 arcsin (\frac{x}{2}) - 18 sin (2 arcsin (\frac{x}{2})))]_{- 2}^{2}

Compute the boundaries:

- 6 π

= - 6 π

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What is the integral from-2 to 2 of (x-3)sqrt(4-x^2) ?
The integral from-2 to 2 of (x-3)sqrt(4-x^2) is -6pi