What is the integral from 0 to 2 of (2x^3-2x^2-4x-1) ?

The integral from 0 to 2 of (2x^3-2x^2-4x-1) is -22/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 2 of (2x^3-2x^2-4x-1)

\int_{0}^{2} (2 x^{3} - 2 x^{2} - 4 x - 1) d x

- \frac{22}{3}

Decimal

- 7.33333 \dots

Solution steps

\int_{0}^{2} 2 x^{3} - 2 x^{2} - 4 x - 1 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}^{2} 2 x^{3} d x - \int_{0}^{2} 2 x^{2} d x - \int_{0}^{2} 4 x d x - \int_{0}^{2} 1 d x

\int_{0}^{2} 2 x^{3} d x = 8

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

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

\int_{0}^{2} 1 d x = 2

= 8 - \frac{16}{3} - 8 - 2

Simplify

8 - \frac{16}{3} - 8 - 2 : - \frac{22}{3}

= - \frac{22}{3}

What is the integral from 0 to 2 of (2x^3-2x^2-4x-1) ?
The integral from 0 to 2 of (2x^3-2x^2-4x-1) is -22/3