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

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

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

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

16 + 4 π

Decimal

28.56637 \dots

Solution steps

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

\int_{- 4}^{0} 4 d x = 16

\int_{- 4}^{0} 16 - x^{2} d x = 4 π

= 16 + 4 π

\int - 5 e^{6 x} d x

\int \frac{1}{x ^{2} 100 - x ^{2}} d x

x \to - \infty lim (sin (x))

arctan (x)

\int \frac{2}{3} x (a^{2} + x^{2})^{\frac{3}{2}} d x

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