What is the integral from-6 to 0 of (4+sqrt(36-x^2)) ?

The integral from-6 to 0 of (4+sqrt(36-x^2)) is 24+9pi

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integral from-6 to 0 of (4+sqrt(36-x^2))

\int_{- 6}^{0} (4 + 36 - x^{2}) d x

24 + 9 π

Decimal

52.27433 \dots

Solution steps

\int_{- 6}^{0} 4 + 36 - 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_{- 6}^{0} 4 d x + \int_{- 6}^{0} 36 - x^{2} d x

\int_{- 6}^{0} 4 d x = 24

\int_{- 6}^{0} 36 - x^{2} d x = 9 π

= 24 + 9 π

What is the integral from-6 to 0 of (4+sqrt(36-x^2)) ?
The integral from-6 to 0 of (4+sqrt(36-x^2)) is 24+9pi