What is the integral from 0 to 2 of 28-(3x^2+2x+12) ?

The integral from 0 to 2 of 28-(3x^2+2x+12) is 20

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 28-(3x^2+2x+12)

\int_{0}^{2} 28 - (3 x^{2} + 2 x + 12) d x

20

Solution steps

\int_{0}^{2} 28 - (3 x^{2} + 2 x + 12) d x

Simplify

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

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

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

\int_{0}^{2} 16 d x = 32

= - 8 - 4 + 32

Simplify

= 20

What is the integral from 0 to 2 of 28-(3x^2+2x+12) ?
The integral from 0 to 2 of 28-(3x^2+2x+12) is 20