What is the integral from-3 to 3 of (1+sqrt(9-x^2)) ?

The integral from-3 to 3 of (1+sqrt(9-x^2)) is 6+(9pi)/2

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integral from-3 to 3 of (1+sqrt(9-x^2))

\int_{- 3}^{3} (1 + 9 - x^{2}) d x

6 + \frac{9 π}{2}

Decimal

20.13716 \dots

Solution steps

\int_{- 3}^{3} 1 + 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}^{3} 1 d x + \int_{- 3}^{3} 9 - x^{2} d x

\int_{- 3}^{3} 1 d x = 6

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

= 6 + \frac{9 π}{2}

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What is the integral from-3 to 3 of (1+sqrt(9-x^2)) ?
The integral from-3 to 3 of (1+sqrt(9-x^2)) is 6+(9pi)/2