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

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

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

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

3 + \frac{9 π}{4}

Decimal Notation

10.06858 \dots

Solution steps

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

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

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

= 3 + \frac{9 π}{4}

\frac{200}{( s + 10 ) ( s + 100 )}

x^{^{''}} - x = 0

\frac{d}{d x} (x^{3} - 2 x^{2} + x - 20)

\int_{0}^{1} x sin (nπ x) d x

\frac{\partial}{\partial y} (y^{4})

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