What is the integral from 0 to 1 of (e^x-1)^4e^x ?

The integral from 0 to 1 of (e^x-1)^4e^x is ((e-1)^5)/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of (e^x-1)^4e^x

\int_{0}^{1} (e^{x} - 1)^{4} e^{x} d x

\frac{( e - 1 ) ^{5}}{5}

Decimal

2.99572 \dots

Solution steps

\int_{0}^{1} (e^{x} - 1)^{4} e^{x} d x

Apply u-substitution

: \int_{0}^{e - 1} u^{4} d u

= \int_{0}^{e - 1} u^{4} d u

Apply the Power Rule

: [\frac{u ^{5}}{5}]_{0}^{e - 1}

= [\frac{u ^{5}}{5}]_{0}^{e - 1}

Compute the boundaries:

\frac{( e - 1 ) ^{5}}{5}

= \frac{( e - 1 ) ^{5}}{5}

\int_{a}^{b} x \frac{1}{b - a} d x

\int_{1}^{6} \frac{9}{x} d x

\int_{0}^{2} x \cdot \frac{x}{2} d x

\int_{0}^{1} 6 y d y

\int_{2}^{5} (x^{2} - 8 x + 12) d x

What is the integral from 0 to 1 of (e^x-1)^4e^x ?
The integral from 0 to 1 of (e^x-1)^4e^x is ((e-1)^5)/5