What is the integral from 1 to 8 of (t^2-4t-12) ?

The integral from 1 to 8 of (t^2-4t-12) is -119/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 8 of (t^2-4t-12)

\int_{1}^{8} (t^{2} - 4 t - 12) d t

- \frac{119}{3}

Decimal

- 39.66666 \dots

Solution steps

\int_{1}^{8} t^{2} - 4 t - 12 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_{1}^{8} t^{2} d t - \int_{1}^{8} 4 t d t - \int_{1}^{8} 12 d t

\int_{1}^{8} t^{2} d t = \frac{511}{3}

\int_{1}^{8} 4 t d t = 126

\int_{1}^{8} 12 d t = 84

= \frac{511}{3} - 126 - 84

Simplify

\frac{511}{3} - 126 - 84 : - \frac{119}{3}

= - \frac{119}{3}

\int_{0}^{1} 1 x + 6 d x

\int_{0}^{t} t sin (t) d t

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

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

\int_{0}^{θ} x (\frac{2 x}{θ ^{2}}) d x

What is the integral from 1 to 8 of (t^2-4t-12) ?
The integral from 1 to 8 of (t^2-4t-12) is -119/3