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

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

\int_{0}^{b} e^{\frac{x}{2}} d x

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

Solution steps

\int_{0}^{b} e^{\frac{x}{2}} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Compute the boundaries:

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

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

\int \frac{x ^{3}}{3 x} d x

x \to 2 lim (\frac{f ( x )}{g ( x )})

\int \frac{( x ^{2} - x + 6 )}{( x ^{3} - 3 x )} d x

x^{2} - 6

(2 + x) (\frac{d y}{d x}) = 3 y

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