What is the integral from 0 to x of e^{2t} ?

The integral from 0 to x of e^{2t} is 1/2 (e^{2x}-1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to x of e^{2t}

\int_{0}^{x} e^{2 t} d t

\frac{1}{2} (e^{2 x} - 1)

Solution steps

\int_{0}^{x} e^{2 t} d t

Apply u-substitution

: \int_{0}^{2 x} e^{u} \frac{1}{2} d u

= \int_{0}^{2 x} e^{u} \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{2 x} e^{u} d u

Use the common integral:

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

= \frac{1}{2} [e^{u}]_{0}^{2 x}

Compute the boundaries:

e^{2 x} - 1

= \frac{1}{2} (e^{2 x} - 1)

\int_{- 3}^{3} - \frac{5}{x ^{3}} d x

\int_{- 1}^{0} \frac{1}{1 - x} d x

\int_{1}^{l n (x)} (t^{2} + 3) d t

\int_{- \infty}^{0} x e^{- 3 x} d x

\int_{0}^{2} x e^{4 - x^{2}} d x

What is the integral from 0 to x of e^{2t} ?
The integral from 0 to x of e^{2t} is 1/2 (e^{2x}-1)