What is the integral from-6 to 0 of (2+sqrt(36-x^2)) ?

The integral from-6 to 0 of (2+sqrt(36-x^2)) is 12+9pi

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integral from-6 to 0 of (2+sqrt(36-x^2))

\int_{- 6}^{0} (2 + 36 - x^{2}) d x

12 + 9 π

Decimal

40.27433 \dots

Solution steps

\int_{- 6}^{0} 2 + 36 - 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_{- 6}^{0} 2 d x + \int_{- 6}^{0} 36 - x^{2} d x

\int_{- 6}^{0} 2 d x = 12

\int_{- 6}^{0} 36 - x^{2} d x = 9 π

= 12 + 9 π

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

y^{^{'}} + 6 y = t + e^{- 5 t}

\int sin^{4} (x) \cdot cos^{4} (x) d x

\int \frac{sec ^{2} ( ln ( x ) )}{2 x} d x

x \to 1 lim (3 x)

What is the integral from-6 to 0 of (2+sqrt(36-x^2)) ?
The integral from-6 to 0 of (2+sqrt(36-x^2)) is 12+9pi