What is the integral from 1 to 2 of e^x-ln(x) ?

The integral from 1 to 2 of e^x-ln(x) is e^2-e-2ln(2)+1

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^x-ln(x)

\int_{1}^{2} e^{x} - ln (x) d x

e^{2} - e - 2 ln (2) + 1

Decimal

4.28447 \dots

Solution steps

\int_{1}^{2} e^{x} - ln (x) 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_{1}^{2} e^{x} d x - \int_{1}^{2} ln (x) d x

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

\int_{1}^{2} ln (x) d x = 2 ln (2) - 1

= e^{2} - e - (2 ln (2) - 1)

Simplify

= e^{2} - e - 2 ln (2) + 1

What is the integral from 1 to 2 of e^x-ln(x) ?
The integral from 1 to 2 of e^x-ln(x) is e^2-e-2ln(2)+1