What is the integral from 1 to ln(8) of integral from 0 to ln(y) of e^{x+y} ?

The integral from 1 to ln(8) of integral from 0 to ln(y) of e^{x+y} is 24ln(2)-16+e

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 1 to ln(8) of integral from 0 to ln(y) of e^{x+y}

\int_{1}^{l n (8)} \int_{0}^{l n (y)} e^{x + y} d x d y

24 ln (2) - 16 + e

Decimal

3.35381 \dots

Solution steps

\int_{1}^{l n (8)} \int_{0}^{l n (y)} e^{x + y} d x d y

\int_{0}^{l n (y)} e^{x + y} d x = e^{y} (y - 1)

= \int_{1}^{l n (8)} e^{y} (y - 1) d y

\int_{1}^{l n (8)} e^{y} (y - 1) d y = 24 ln (2) - 16 + e

= 24 ln (2) - 16 + e

\int \frac{x}{( x ^{2} + 1 ) ^{\frac{1}{2}}} d x

a re a x^{4} - 9 x^{2}, x^{2} - 9, 0, 3

\int \frac{x}{x ^{2} + 64} d x

\frac{d y}{d x} = 7 y^{2} sin (x)

\int \frac{e ^{x}}{2 + e ^{x}} d x

What is the integral from 1 to ln(8) of integral from 0 to ln(y) of e^{x+y} ?
The integral from 1 to ln(8) of integral from 0 to ln(y) of e^{x+y} is 24ln(2)-16+e