What is the integral from-8 to 0 of (2+sqrt(64-x^2)) ?

The integral from-8 to 0 of (2+sqrt(64-x^2)) is 16+16pi

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integral from-8 to 0 of (2+sqrt(64-x^2))

\int_{- 8}^{0} (2 + 64 - x^{2}) d x

16 + 16 π

Decimal

66.26548 \dots

Solution steps

\int_{- 8}^{0} 2 + 64 - 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_{- 8}^{0} 2 d x + \int_{- 8}^{0} 64 - x^{2} d x

\int_{- 8}^{0} 2 d x = 16

\int_{- 8}^{0} 64 - x^{2} d x = 16 π

= 16 + 16 π

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What is the integral from-8 to 0 of (2+sqrt(64-x^2)) ?
The integral from-8 to 0 of (2+sqrt(64-x^2)) is 16+16pi