What is the integral from 0 to 8 of e^{x/2} ?

The integral from 0 to 8 of e^{x/2} is 2(e^4-1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 8 of e^{x/2}

\int_{0}^{8} e^{\frac{x}{2}} d x

2 (e^{4} - 1)

Decimal Notation

107.19630 \dots

Solution steps

\int_{0}^{8} e^{\frac{x}{2}} d x

Apply u-substitution

: \int_{0}^{4} e^{u} \cdot 2 d u

= \int_{0}^{4} 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}^{4} e^{u} d u

Use the common integral:

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

= 2 [e^{u}]_{0}^{4}

Compute the boundaries:

e^{4} - 1

= 2 (e^{4} - 1)

f (x) = x + \frac{49}{x}

\frac{\partial}{\partial x} (x^{3})

y^{^{'}} + 10 y = 20

n = 1 \sum \infty n + (\frac{7}{8})^{n}

y = x^{3} - x^{2} + x - 1, (1, 0)

What is the integral from 0 to 8 of e^{x/2} ?
The integral from 0 to 8 of e^{x/2} is 2(e^4-1)