What is the integral from 0 to 1 of e^{15x-14} ?

The integral from 0 to 1 of e^{15x-14} is (e^{15}-1)/(15e^{14)}

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^{15x-14}

\int_{0}^{1} e^{15 x - 14} d x

\frac{e ^{15} - 1}{15 e ^{14}}

Decimal

0.18121 \dots

Solution steps

\int_{0}^{1} e^{15 x - 14} d x

Apply u-substitution

= \int_{- 14}^{1} e^{u} \frac{1}{15} d u

Take the constant out:

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

= \frac{1}{15} \cdot \int_{- 14}^{1} e^{u} d u

Use the common integral:

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

= \frac{1}{15} [e^{u}]_{- 14}^{1}

Compute the boundaries:

e - \frac{1}{e ^{14}}

= \frac{1}{15} (e - \frac{1}{e ^{14}})

Simplify

= \frac{e ^{15} - 1}{15 e ^{14}}

What is the integral from 0 to 1 of e^{15x-14} ?
The integral from 0 to 1 of e^{15x-14} is (e^{15}-1)/(15e^{14)}