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

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

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

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

\frac{9}{2} + \frac{9 π}{4}

Decimal Notation

11.56858 \dots

Solution steps

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

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

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

= \frac{9}{2} + \frac{9 π}{4}

\int_{1}^{2} cos (x^{2}) d x

\frac{\partial}{\partial y} (x^{2} + x y)

\frac{d}{d x} (\frac{9 x}{x})

\int \frac{2 - x}{x} d x

\frac{d}{d x} (cos^{2} (1 - x^{2}))

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