What is the integral from 0 to 1 of e^{tx} ?

The integral from 0 to 1 of e^{tx} is (e^t-1)/t

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^{tx}

\int_{0}^{1} e^{t x} d x

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

Solution steps

\int_{0}^{1} e^{t x} d x

Apply u-substitution

: \int_{0}^{t} \frac{e ^{u}}{t} d u

= \int_{0}^{t} \frac{e ^{u}}{t} d u

Take the constant out:

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

= \frac{1}{t} \cdot \int_{0}^{t} e^{u} d u

Use the common integral:

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

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

Simplify

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

= \frac{[ e ^{u} ] _{0}^{t}}{t}

Compute the boundaries:

e^{t} - 1

= \frac{e ^{t} - 1}{t}

\int_{0}^{1} 4 x + 5 d x

\int_{1}^{2} \frac{x + 1}{x} d x

\int_{- \frac{π}{3}}^{0} sin (x) d x

\int_{0}^{x} e^{x - t} y d t

\int_{0}^{t} x e^{x} d x

What is the integral from 0 to 1 of e^{tx} ?
The integral from 0 to 1 of e^{tx} is (e^t-1)/t