What is the integral from-8 to 0 of 2+sqrt(64-x^2) ?

The integral from-8 to 0 of 2+sqrt(64-x^2) is 16+16pi

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integral from-8 to 0 of 2+sqrt(64-x^2)

\int_{- 8}^{0} 2 + 64 - x^{2} d x

16 + 16 π

Decimal Notation

66.26548 \dots

Solution steps

\int_{- 8}^{0} 2 + 64 - 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_{- 8}^{0} 2 d x + \int_{- 8}^{0} 64 - x^{2} d x

\int_{- 8}^{0} 2 d x = 16

\int_{- 8}^{0} 64 - x^{2} d x = 16 π

= 16 + 16 π

y = \frac{2}{x - 2}, (4, 1)

\frac{d}{d t} (t^{2} + t)

f (x) = \frac{3 cos ( x )}{3 x ^{2} + 3 x - 1}

\frac{d}{d x} (ln (c x))

What is the integral from-8 to 0 of 2+sqrt(64-x^2) ?
The integral from-8 to 0 of 2+sqrt(64-x^2) is 16+16pi