What is the integral from 1 to 13 of (x-1)/(72) ?

The integral from 1 to 13 of (x-1)/(72) is 1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 1 to 13 of (x-1)/(72)

\int_{1}^{13} \frac{x - 1}{72} d x

1

Solution steps

\int_{1}^{13} \frac{x - 1}{72} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{72} \cdot \int_{1}^{13} 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

= \frac{1}{72} (\int_{1}^{13} x d x - \int_{1}^{13} 1 d x)

\int_{1}^{13} x d x = 84

\int_{1}^{13} 1 d x = 12

= \frac{1}{72} (84 - 12)

Simplify

\frac{1}{72} (84 - 12) : 1

= 1

What is the integral from 1 to 13 of (x-1)/(72) ?
The integral from 1 to 13 of (x-1)/(72) is 1