What is the integral from-2 to 2 of (x+9)sqrt(4-x^2) ?

The integral from-2 to 2 of (x+9)sqrt(4-x^2) is 18pi

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integral from-2 to 2 of (x+9)sqrt(4-x^2)

\int_{- 2}^{2} (x + 9) 4 - x^{2} d x

18 π

Decimal Notation

56.54866 \dots

Solution steps

\int_{- 2}^{2} (x + 9) 4 - x^{2} d x

Expand

(x + 9) 4 - x^{2} : x 4 - x^{2} + 9 4 - x^{2}

= \int_{- 2}^{2} x 4 - x^{2} + 9 4 - x^{2} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 2}^{2} x 4 - x^{2} d x + \int_{- 2}^{2} 9 4 - x^{2} d x

\int_{- 2}^{2} x 4 - x^{2} d x = 0

\int_{- 2}^{2} 9 4 - x^{2} d x = 18 π

= 0 + 18 π

Simplify

= 18 π

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What is the integral from-2 to 2 of (x+9)sqrt(4-x^2) ?
The integral from-2 to 2 of (x+9)sqrt(4-x^2) is 18pi