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

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

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

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

8 π + 16

Decimal

41.13274 \dots

Solution steps

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

\int_{- 4}^{4} 16 - x^{2} d x = 8 π

\int_{- 4}^{4} 2 d x = 16

= 8 π + 16

\int_{3}^{4} - x^{3} + 7 x^{2} - 12 x d x

\int_{0}^{\infty} x ln (x) d x

\int_{0}^{1} (8 x^{e} + 7 e^{x}) d x

\int_{- 5}^{0} 0 d x

\int_{1}^{2} ln (1 + x) d x

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