What is the integral from 0 to 1 of e^{2x} ?

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

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^{2x}

\int_{0}^{1} e^{2 x} d x

\frac{e ^{2} - 1}{2}

Decimal Notation

3.19452 \dots

Solution steps

\int_{0}^{1} e^{2 x} d x

Apply u-substitution

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

= \int_{0}^{2} 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} e^{u} d u

Use the common integral:

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

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

Compute the boundaries:

e^{2} - 1

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

Simplify

= \frac{e ^{2} - 1}{2}

f (x) = x^{3} - 2 x - 2, at x = - 1

f (t) = \frac{9}{t ^{2}}

x (y^{^{'}}) = 4 y

x^{3} + 2

\int 4 ln (x^{2} - 1) d x

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