What is the integral from 0 to 2 of c(4x-2x^2) ?

The integral from 0 to 2 of c(4x-2x^2) is 8/3 c

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 2 of c(4x-2x^2)

\int_{0}^{2} c (4 x - 2 x^{2}) d x

\frac{8}{3} c

Solution steps

\int_{0}^{2} c (4 x - 2 x^{2}) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= c \cdot \int_{0}^{2} 4 x - 2 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

= c (\int_{0}^{2} 4 x d x - \int_{0}^{2} 2 x^{2} d x)

\int_{0}^{2} 4 x d x = 8

\int_{0}^{2} 2 x^{2} d x = \frac{16}{3}

= c (8 - \frac{16}{3})

Join

8 - \frac{16}{3} : \frac{8}{3}

= \frac{8}{3} c

\int_{\frac{x}{4}}^{\frac{x}{2}} sin (x) d x

\int_{2}^{3} \frac{x}{x - 2} d x

\int_{8}^{8} x 3^{x} d x

\int_{- 3}^{4} [y - (y^{2} - 12)] d y

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

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