What is the integral from-4 to 4 of (6+sqrt(16-x^2)) ?

The integral from-4 to 4 of (6+sqrt(16-x^2)) is 48+8pi

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integral from-4 to 4 of (6+sqrt(16-x^2))

\int_{- 4}^{4} (6 + 16 - x^{2}) d x

48 + 8 π

Decimal

73.13274 \dots

Solution steps

\int_{- 4}^{4} 6 + 16 - 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_{- 4}^{4} 6 d x + \int_{- 4}^{4} 16 - x^{2} d x

\int_{- 4}^{4} 6 d x = 48

\int_{- 4}^{4} 16 - x^{2} d x = 8 π

= 48 + 8 π

What is the integral from-4 to 4 of (6+sqrt(16-x^2)) ?
The integral from-4 to 4 of (6+sqrt(16-x^2)) is 48+8pi