What is the integral of e^{0.5t} ?

The integral of e^{0.5t} is 2e^{0.5t}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of e^{0.5t}

\int e^{0.5 t} d t

2 e^{0.5 t} + C

Solution steps

\int e^{0.5 t} d t

Apply u-substitution

= \int e^{u} \frac{1}{0.5} d u

Take the constant out:

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

= \frac{1}{0.5} \cdot \int e^{u} d u

Simplify

= 2 \cdot \int e^{u} d u

Use the common integral:

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

= 2 e^{u}

Substitute back

u = 0.5 t

= 2 e^{0.5 t}

Add a constant to the solution

= 2 e^{0.5 t} + C

x \to 2 lim (\frac{( 3 x - 6 )}{1 - 4 x - 7})

\frac{\partial}{\partial y} (x^{2} y^{3})

\int 8 cos (x) d x

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} - 3 y = 2 x - 17

y = - 2 x^{2} - x + 6