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{d ^{2}}{d x ^{2}} (\frac{1}{x ^{4}})

n = 1 \sum \infty 4^{- 2 n}

y^{^{'}} = y \cdot (x + 1)

x + 2 sin (x)

f (x) = \frac{x}{x ^{2} + 9}

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