What is the integral from 0 to 3 of 1+sqrt(9-x^2) ?

The integral from 0 to 3 of 1+sqrt(9-x^2) is 3+(9pi)/4

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integral from 0 to 3 of 1+sqrt(9-x^2)

\int_{0}^{3} 1 + 9 - x^{2} d x

3 + \frac{9 π}{4}

Decimal

10.06858 \dots

Solution steps

\int_{0}^{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_{0}^{3} 1 d x + \int_{0}^{3} 9 - x^{2} d x

\int_{0}^{3} 1 d x = 3

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

= 3 + \frac{9 π}{4}

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What is the integral from 0 to 3 of 1+sqrt(9-x^2) ?
The integral from 0 to 3 of 1+sqrt(9-x^2) is 3+(9pi)/4