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

The integral from-4 to 4 of 4+sqrt(16-x^2) is 32+8pi

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

32 + 8 π

Decimal

57.13274 \dots

Solution steps

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

\int_{- 4}^{4} 4 d x = 32

\int_{- 4}^{4} 16 - x^{2} d x = 8 π

= 32 + 8 π

\int_{0}^{2} (6 x - 6) (4 x^{2} + 8) d x

\int_{3}^{9} 1 - \frac{9}{x} d x

\int_{0}^{1} \frac{2}{x ( x + 1 )} d x

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

\int_{2}^{3} x (x^{2} - 4)^{3} d x

What is the integral from-4 to 4 of 4+sqrt(16-x^2) ?
The integral from-4 to 4 of 4+sqrt(16-x^2) is 32+8pi