What is the integral from 0 to 1 of (e^x+x^e) ?

The integral from 0 to 1 of (e^x+x^e) is e-1+1/(e+1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 1 of (e^x+x^e)

\int_{0}^{1} (e^{x} + x^{e}) d x

e - 1 + \frac{1}{e + 1}

Decimal

1.98722 \dots

Solution steps

\int_{0}^{1} e^{x} + x^{e} 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_{0}^{1} e^{x} d x + \int_{0}^{1} x^{e} d x

\int_{0}^{1} e^{x} d x = e - 1

\int_{0}^{1} x^{e} d x = \frac{1}{e + 1}

= e - 1 + \frac{1}{e + 1}

What is the integral from 0 to 1 of (e^x+x^e) ?
The integral from 0 to 1 of (e^x+x^e) is e-1+1/(e+1)