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

The integral from-4 to 4 of (1+sqrt(16-x^2)) is 8+8pi

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

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

8 + 8 π

Decimal

33.13274 \dots

Solution steps

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

\int_{- 4}^{4} 1 d x = 8

\int_{- 4}^{4} 16 - x^{2} d x = 8 π

= 8 + 8 π

\frac{d}{d x} (\frac{8 x}{( x ^{2} + 4 ) ^{2}})

\int \frac{x ^{5}}{x ^{6} + 7} d x

\frac{\partial}{\partial x} (\frac{4}{3} π r^{3})

(x^{2} - 3 x + 3) 25 - x^{2}

\int \frac{x ^{2}}{x + 8} d x

What is the integral from-4 to 4 of (1+sqrt(16-x^2)) ?
The integral from-4 to 4 of (1+sqrt(16-x^2)) is 8+8pi