What is the integral from 0 to 4 of x/3+sqrt(16-x^2) ?

The integral from 0 to 4 of x/3+sqrt(16-x^2) is 8/3+4pi

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 4 of x/3+sqrt(16-x^2)

\int_{0}^{4} \frac{x}{3} + 16 - x^{2} d x

\frac{8}{3} + 4 π

Decimal

15.23303 \dots

Solution steps

\int_{0}^{4} \frac{x}{3} + 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_{0}^{4} \frac{x}{3} d x + \int_{0}^{4} 16 - x^{2} d x

\int_{0}^{4} \frac{x}{3} d x = \frac{8}{3}

\int_{0}^{4} 16 - x^{2} d x = 4 π

= \frac{8}{3} + 4 π

\int_{0}^{1} (1 + 5 x)^{3} d x

\int_{0}^{1} 3 x^{2} \cdot e^{x^{3}} d x

\int_{0}^{\frac{π}{4}} (2 cos^{2} (x)) d x

\int_{- π}^{0} x cos ((2 n + 1) x) d x

\int_{7}^{2} x d x

What is the integral from 0 to 4 of x/3+sqrt(16-x^2) ?
The integral from 0 to 4 of x/3+sqrt(16-x^2) is 8/3+4pi