What is the integral from 0 to 1 of e^{7x-6} ?

The integral from 0 to 1 of e^{7x-6} is (e^7-1)/(7e^6)

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^{7x-6}

\int_{0}^{1} e^{7 x - 6} d x

\frac{e ^{7} - 1}{7 e ^{6}}

Decimal

0.38797 \dots

Solution steps

\int_{0}^{1} e^{7 x - 6} d x

Apply u-substitution

= \int_{- 6}^{1} e^{u} \frac{1}{7} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{7} \cdot \int_{- 6}^{1} e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= \frac{1}{7} [e^{u}]_{- 6}^{1}

Compute the boundaries:

e - \frac{1}{e ^{6}}

= \frac{1}{7} (e - \frac{1}{e ^{6}})

Simplify

= \frac{e ^{7} - 1}{7 e ^{6}}

What is the integral from 0 to 1 of e^{7x-6} ?
The integral from 0 to 1 of e^{7x-6} is (e^7-1)/(7e^6)