What is the integral from 0 to 2 of (3x^2-6x+1) ?

The integral from 0 to 2 of (3x^2-6x+1) is -2

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^2-6x+1)

\int_{0}^{2} (3 x^{2} - 6 x + 1) d x

- 2

Solution steps

\int_{0}^{2} 3 x^{2} - 6 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} 6 x d x + \int_{0}^{2} 1 d x

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

\int_{0}^{2} 6 x d x = 12

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

= 8 - 12 + 2

Simplify

= - 2

y = x^{2} - 2 x - 24 x

y^{^{'}} = - \frac{x}{y} + \frac{( x ^{2} + y ^{2} )}{y}, y (0) = - 1

\int cos (x) (3 + 5 sin^{2} (x)) d x

\int \frac{2 x + 1}{( x + 7 ) ^{2}} d x

\frac{d}{d x} (arcsin (x + y))

What is the integral from 0 to 2 of (3x^2-6x+1) ?
The integral from 0 to 2 of (3x^2-6x+1) is -2