What is the integral from-4 to 0 of 2+sqrt(16-x^2) ?

The integral from-4 to 0 of 2+sqrt(16-x^2) is 8+4pi

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integral from-4 to 0 of 2+sqrt(16-x^2)

\int_{- 4}^{0} 2 + 16 - x^{2} d x

8 + 4 π

Decimal Notation

20.56637 \dots

Solution steps

\int_{- 4}^{0} 2 + 16 - 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_{- 4}^{0} 2 d x + \int_{- 4}^{0} 16 - x^{2} d x

\int_{- 4}^{0} 2 d x = 8

\int_{- 4}^{0} 16 - x^{2} d x = 4 π

= 8 + 4 π

\int_{- 1}^{2} x - x^{2} + 2 d x

\frac{\partial}{\partial x} (2 x^{2} + 4 y^{2})

f (t) = e^{5 t}

\int sin^{2} (1) d x

\frac{\partial}{\partial y} (- \frac{1}{x ^{2} y})

What is the integral from-4 to 0 of 2+sqrt(16-x^2) ?
The integral from-4 to 0 of 2+sqrt(16-x^2) is 8+4pi