What is the integral from 1 to 2 of (e^t+4t) ?

The integral from 1 to 2 of (e^t+4t) is e^2-e+6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 1 to 2 of (e^t+4t)

\int_{1}^{2} (e^{t} + 4 t) d t

e^{2} - e + 6

Decimal

10.67077 \dots

Solution steps

\int_{1}^{2} e^{t} + 4 t 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}^{2} e^{t} d t + \int_{1}^{2} 4 t d t

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

\int_{1}^{2} 4 t d t = 6

= e^{2} - e + 6

What is the integral from 1 to 2 of (e^t+4t) ?
The integral from 1 to 2 of (e^t+4t) is e^2-e+6