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

The integral from-2 to-1 of (4x^2+3x-1) is 23/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

\frac{23}{6}

Decimal

3.83333 \dots

Solution steps

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

\int_{- 2}^{- 1} 4 x^{2} d x = \frac{28}{3}

\int_{- 2}^{- 1} 3 x d x = - \frac{9}{2}

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

= \frac{28}{3} - \frac{9}{2} - 1

Simplify

\frac{28}{3} - \frac{9}{2} - 1 : \frac{23}{6}

= \frac{23}{6}

What is the integral from-2 to-1 of (4x^2+3x-1) ?
The integral from-2 to-1 of (4x^2+3x-1) is 23/6