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

The integral from 0 to 2 of (3x^2-x+1) is 8

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 (3x^2-x+1)

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

8

Solution steps

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

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

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

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

= 8 - 2 + 2

Simplify

= 8

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

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

\int_{0}^{5} y^{2} d y

\int_{1}^{e} \frac{t ^{3} + 1}{t} d t

\int_{0}^{π} cos^{2} (x) \cdot sin (x) d x

What is the integral from 0 to 2 of (3x^2-x+1) ?
The integral from 0 to 2 of (3x^2-x+1) is 8