What is the integral from-2 to 2 of (x-1)sqrt(4-x^2) ?

The integral from-2 to 2 of (x-1)sqrt(4-x^2) is -2pi

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integral from-2 to 2 of (x-1)sqrt(4-x^2)

\int_{- 2}^{2} (x - 1) 4 - x^{2} d x

- 2 π

Decimal

- 6.28318 \dots

Solution steps

\int_{- 2}^{2} (x - 1) 4 - x^{2} d x

Multiply by the conjugate of

x - 1 : \frac{x ^{2} - 1}{x + 1}

= \int_{- 2}^{2} \frac{x ^{2} - 1}{x + 1} 4 - x^{2} d x

Find undefined (singularity) points:

x = - 1

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{- 2}^{- 1} \frac{x ^{2} - 1}{x + 1} 4 - x^{2} d x + \int_{- 1}^{2} \frac{x ^{2} - 1}{x + 1} 4 - x^{2} d x

\int_{- 2}^{- 1} \frac{x ^{2} - 1}{x + 1} 4 - x^{2} d x = - 3 - 2 (\frac{π}{3} - \frac{3}{4})

\int_{- 1}^{2} \frac{x ^{2} - 1}{x + 1} 4 - x^{2} d x = 3 - 2 (\frac{2 π}{3} + \frac{3}{4})

= - 3 - 2 (\frac{π}{3} - \frac{3}{4}) + 3 - 2 (\frac{2 π}{3} + \frac{3}{4})

Simplify

- 3 - 2 (\frac{π}{3} - \frac{3}{4}) + 3 - 2 (\frac{2 π}{3} + \frac{3}{4}) : - 2 π

= - 2 π

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What is the integral from-2 to 2 of (x-1)sqrt(4-x^2) ?
The integral from-2 to 2 of (x-1)sqrt(4-x^2) is -2pi