What is the integral from-8 to 0 of (1+sqrt(64-x^2)) ?

The integral from-8 to 0 of (1+sqrt(64-x^2)) is 8+16pi

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integral from-8 to 0 of (1+sqrt(64-x^2))

\int_{- 8}^{0} (1 + 64 - x^{2}) d x

8 + 16 π

Decimal Notation

58.26548 \dots

Solution steps

\int_{- 8}^{0} 1 + 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} 1 d x + \int_{- 8}^{0} 64 - x^{2} d x

\int_{- 8}^{0} 1 d x = 8

\int_{- 8}^{0} 64 - x^{2} d x = 16 π

= 8 + 16 π

f (x) = 1 + x^{3}

n = 0 \sum \infty n^{n}

\frac{\partial}{\partial x} (4 x y + 6 y^{2})

f (x) = (x - 1)^{3}, at x = 2

\int \frac{1}{x ( 2 a - x )} d x

What is the integral from-8 to 0 of (1+sqrt(64-x^2)) ?
The integral from-8 to 0 of (1+sqrt(64-x^2)) is 8+16pi