What is the integral from-3 to 0 of (4+sqrt(9-x^2)) ?

The integral from-3 to 0 of (4+sqrt(9-x^2)) is 12+(9pi)/4

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integral from-3 to 0 of (4+sqrt(9-x^2))

\int_{- 3}^{0} (4 + 9 - x^{2}) d x

12 + \frac{9 π}{4}

Decimal Notation

19.06858 \dots

Solution steps

\int_{- 3}^{0} 4 + 9 - 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_{- 3}^{0} 4 d x + \int_{- 3}^{0} 9 - x^{2} d x

\int_{- 3}^{0} 4 d x = 12

\int_{- 3}^{0} 9 - x^{2} d x = \frac{9 π}{4}

= 12 + \frac{9 π}{4}

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What is the integral from-3 to 0 of (4+sqrt(9-x^2)) ?
The integral from-3 to 0 of (4+sqrt(9-x^2)) is 12+(9pi)/4