What is the integral from-2 to 0 of (3t^3-t^2-9t-1) ?

The integral from-2 to 0 of (3t^3-t^2-9t-1) is 4/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from-2 to 0 of (3t^3-t^2-9t-1)

\int_{- 2}^{0} (3 t^{3} - t^{2} - 9 t - 1) d t

\frac{4}{3}

Decimal

1.33333 \dots

Solution steps

\int_{- 2}^{0} 3 t^{3} - t^{2} - 9 t - 1 d t

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}^{0} 3 t^{3} d t - \int_{- 2}^{0} t^{2} d t - \int_{- 2}^{0} 9 t d t - \int_{- 2}^{0} 1 d t

\int_{- 2}^{0} 3 t^{3} d t = - 12

\int_{- 2}^{0} t^{2} d t = \frac{8}{3}

\int_{- 2}^{0} 9 t d t = - 18

\int_{- 2}^{0} 1 d t = 2

= - 12 - \frac{8}{3} - (- 18) - 2

Simplify

- 12 - \frac{8}{3} - (- 18) - 2 : \frac{4}{3}

= \frac{4}{3}

What is the integral from-2 to 0 of (3t^3-t^2-9t-1) ?
The integral from-2 to 0 of (3t^3-t^2-9t-1) is 4/3