What is the integral from-4 to 0 of (1+sqrt(16-x^2)) ?

The integral from-4 to 0 of (1+sqrt(16-x^2)) is 4+4pi

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integral from-4 to 0 of (1+sqrt(16-x^2))

\int_{- 4}^{0} (1 + 16 - x^{2}) d x

4 + 4 π

Decimal

16.56637 \dots

Solution steps

\int_{- 4}^{0} 1 + 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} 1 d x + \int_{- 4}^{0} 16 - x^{2} d x

\int_{- 4}^{0} 1 d x = 4

\int_{- 4}^{0} 16 - x^{2} d x = 4 π

= 4 + 4 π

\int t^{2} 13 t - 1 d t

f (x) = (\frac{4}{x}) + (\frac{1}{5 x ^{3} + 2 x})

\int_{0}^{\frac{π}{2}} sec^{2} (\frac{x}{2}) d x

x y^{^{'}} - 2 y = 0

y^{^{'}} + y = 4 x

What is the integral from-4 to 0 of (1+sqrt(16-x^2)) ?
The integral from-4 to 0 of (1+sqrt(16-x^2)) is 4+4pi